id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-7200 | Exclusive Or is Negation of Biconditional | Exclusive or is equivalent to the negation of the biconditional:
:$p \oplus q \dashv \vdash \neg \paren {p \iff q}$ | {{begin-eqn}}
{{eqn | l = p \oplus q
| o = \dashv \vdash
| r = \paren {p \lor q} \land \neg \paren {p \land q}
| c = {{Defof|Exclusive Or}}
}}
{{eqn | o = \dashv \vdash
| r = \neg \paren {p \iff q}
| c = Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction
}}
{{end-eq... | [[Definition:Exclusive Or|Exclusive or]] is [[Definition:Logically Equivalent|equivalent]] to the [[Definition:Logical Not|negation]] of the [[Definition:Biconditional|biconditional]]:
:$p \oplus q \dashv \vdash \neg \paren {p \iff q}$ | {{begin-eqn}}
{{eqn | l = p \oplus q
| o = \dashv \vdash
| r = \paren {p \lor q} \land \neg \paren {p \land q}
| c = {{Defof|Exclusive Or}}
}}
{{eqn | o = \dashv \vdash
| r = \neg \paren {p \iff q}
| c = [[Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction]]
}}
{{en... | Exclusive Or is Negation of Biconditional | https://proofwiki.org/wiki/Exclusive_Or_is_Negation_of_Biconditional | https://proofwiki.org/wiki/Exclusive_Or_is_Negation_of_Biconditional | [
"Exclusive Or",
"Biconditional"
] | [
"Definition:Exclusive Or",
"Definition:Logical Equivalence",
"Definition:Logical Not",
"Definition:Biconditional"
] | [
"Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction"
] |
proofwiki-7201 | Exclusive Or as Disjunction of Conjunctions | :$p \oplus q \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$ | {{begin-eqn}}
{{eqn | l = p \oplus q
| o = \dashv \vdash
| r = \neg \left ({p \iff q}\right)
| c = Exclusive Or is Negation of Biconditional
}}
{{eqn | o = \dashv \vdash
| r = \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)
| c = Non-Equivalence as Disjunction of Conjuncti... | :$p \oplus q \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$ | {{begin-eqn}}
{{eqn | l = p \oplus q
| o = \dashv \vdash
| r = \neg \left ({p \iff q}\right)
| c = [[Exclusive Or is Negation of Biconditional]]
}}
{{eqn | o = \dashv \vdash
| r = \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)
| c = [[Non-Equivalence as Disjunction of Con... | Exclusive Or as Disjunction of Conjunctions/Proof 1 | https://proofwiki.org/wiki/Exclusive_Or_as_Disjunction_of_Conjunctions | https://proofwiki.org/wiki/Exclusive_Or_as_Disjunction_of_Conjunctions/Proof_1 | [
"Exclusive Or as Disjunction of Conjunctions",
"Exclusive Or",
"Disjunction",
"Conjunction"
] | [] | [
"Exclusive Or is Negation of Biconditional",
"Non-Equivalence as Disjunction of Conjunctions/Formulation 1"
] |
proofwiki-7202 | Exclusive Or as Disjunction of Conjunctions | :$p \oplus q \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin {array} {|ccc||ccccccccc|} \hline
p & \oplus & q & (\neg & p & \land & q) & \lor & (p & \land & \neg & q) \\
\hline
\F & \F & \F & \T & \F & \F & \F & \F... | :$p \oplus q \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin {array} {|ccc||ccccccccc|} \h... | Exclusive Or as Disjunction of Conjunctions/Proof by Truth Table | https://proofwiki.org/wiki/Exclusive_Or_as_Disjunction_of_Conjunctions | https://proofwiki.org/wiki/Exclusive_Or_as_Disjunction_of_Conjunctions/Proof_by_Truth_Table | [
"Exclusive Or as Disjunction of Conjunctions",
"Exclusive Or",
"Disjunction",
"Conjunction"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7203 | Non-Equivalence as Disjunction of Negated Conditionals | :$\neg \paren {p \iff q} \dashv \vdash \neg \paren {p \implies q} \lor \neg \paren {q \implies p}$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \neg \paren {p \implies q} \lor \neg \paren {q \implies p} }}
{{Premise|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }|1
|Rule of Material Equivalence}}
{{DeMorgan|3|1|\neg \paren {p \implies q} \lor \neg \pa... | :$\neg \paren {p \iff q} \dashv \vdash \neg \paren {p \implies q} \lor \neg \paren {q \implies p}$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \neg \paren {p \implies q} \lor \neg \paren {q \implies p} }}
{{Premise|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }|1
|[[Rule of Material Equivalence]]}}
{{DeMorgan|3|1|\neg \paren {p \implies q} \lor \neg... | Non-Equivalence as Disjunction of Negated Conditionals/Proof 1 | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Negated_Conditionals | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Negated_Conditionals/Proof_1 | [
"Non-Equivalence as Disjunction of Negated Conditionals",
"Biconditional",
"Disjunction",
"Logical Negation",
"Conditional"
] | [] | [
"Rule of Material Equivalence",
"Rule of Material Equivalence"
] |
proofwiki-7204 | Non-Equivalence as Disjunction of Negated Conditionals | :$\neg \paren {p \iff q} \dashv \vdash \neg \paren {p \implies q} \lor \neg \paren {q \implies p}$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|cccc||ccccccccc|} \hline
\neg & (p & \iff & q) & \neg & (p & \implies & q) & \lor & \neg & (q & \implies & p) \\
\hline
\F & \F & \T & \F & \F & \F & \T &... | :$\neg \paren {p \iff q} \dashv \vdash \neg \paren {p \implies q} \lor \neg \paren {q \implies p}$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|cccc||ccccccccc|} \hline
\neg... | Non-Equivalence as Disjunction of Negated Conditionals/Proof by Truth Table | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Negated_Conditionals | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Negated_Conditionals/Proof_by_Truth_Table | [
"Non-Equivalence as Disjunction of Negated Conditionals",
"Biconditional",
"Disjunction",
"Logical Negation",
"Conditional"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7205 | Conjunction of Disjunction with Negation is Conjunction with Negation | :$\paren {p \lor q} \land \neg q \dashv \vdash p \land \neg q$ | {{BeginTableau|p \land \neg q \vdash \paren {p \lor q} \land \neg q}}
{{Premise|1|p \land \neg q}}
{{Simplification|2|1|p|1|1}}
{{Addition|3|1|p \lor q|2|1}}
{{Simplification|4|1|\neg q|1|2}}
{{Conjunction|5|1|\paren {p \lor q} \land \neg q|3|4}}
{{EndTableau|lemma}}
and its converse:
{{BeginTableau|\paren {p \lor q} \... | :$\paren {p \lor q} \land \neg q \dashv \vdash p \land \neg q$ | {{BeginTableau|p \land \neg q \vdash \paren {p \lor q} \land \neg q}}
{{Premise|1|p \land \neg q}}
{{Simplification|2|1|p|1|1}}
{{Addition|3|1|p \lor q|2|1}}
{{Simplification|4|1|\neg q|1|2}}
{{Conjunction|5|1|\paren {p \lor q} \land \neg q|3|4}}
{{EndTableau|lemma}}
and its converse:
{{BeginTableau|\paren {p \lor q... | Conjunction of Disjunction with Negation is Conjunction with Negation | https://proofwiki.org/wiki/Conjunction_of_Disjunction_with_Negation_is_Conjunction_with_Negation | https://proofwiki.org/wiki/Conjunction_of_Disjunction_with_Negation_is_Conjunction_with_Negation | [
"Conjunction",
"Disjunction",
"Logical Negation"
] | [] | [
"Rule of Distribution/Conjunction Distributes over Disjunction",
"Modus Tollendo Ponens",
"Category:Conjunction",
"Category:Disjunction",
"Category:Logical Negation"
] |
proofwiki-7206 | Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \lor q} \land \neg \paren {p \land q}$
That is, negation of the biconditional means the same thing as '''either-or but not both''', that is, exclusive or. | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {p \lor q} \land \neg \paren {p \land q} }}
{{Premise|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\paren {\neg p \land q} \lor \paren {p \land \neg q}|1
|Non-Equivalence as Disjunction of Conjunctions: Formulation 1
}}
{{Commutation|3|1|\paren {p \land \neg q} \l... | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \lor q} \land \neg \paren {p \land q}$
That is, [[Definition:Logical Not|negation]] of the [[Definition:Biconditional|biconditional]] means the same thing as '''either-or but not both''', that is, [[Definition:Exclusive Or|exclusive or]]. | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {p \lor q} \land \neg \paren {p \land q} }}
{{Premise|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\paren {\neg p \land q} \lor \paren {p \land \neg q}|1
|[[Non-Equivalence as Disjunction of Conjunctions/Formulation 1|Non-Equivalence as Disjunction of Conjunctions... | Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction/Proof 1 | https://proofwiki.org/wiki/Non-Equivalence_as_Conjunction_of_Disjunction_with_Negation_of_Conjunction | https://proofwiki.org/wiki/Non-Equivalence_as_Conjunction_of_Disjunction_with_Negation_of_Conjunction/Proof_1 | [
"Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction",
"Conjunction",
"Disjunction",
"Logical Negation",
"Biconditional"
] | [
"Definition:Logical Not",
"Definition:Biconditional",
"Definition:Exclusive Or"
] | [
"Non-Equivalence as Disjunction of Conjunctions/Formulation 1",
"Conjunction of Disjunction with Negation is Conjunction with Negation",
"Rule of Distribution/Conjunction Distributes over Disjunction",
"Rule of Distribution/Conjunction Distributes over Disjunction",
"Conjunction of Disjunction with Negation... |
proofwiki-7207 | Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \lor q} \land \neg \paren {p \land q}$
That is, negation of the biconditional means the same thing as '''either-or but not both''', that is, exclusive or. | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|cccc||cccccccc|} \hline
\neg & (p & \iff & q) & (p & \lor & q) & \land & \neg & (p & \land & q) \\
\hline
\F & \F & \T & \F & \F & \F & \F & \F & \T & \F ... | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \lor q} \land \neg \paren {p \land q}$
That is, [[Definition:Logical Not|negation]] of the [[Definition:Biconditional|biconditional]] means the same thing as '''either-or but not both''', that is, [[Definition:Exclusive Or|exclusive or]]. | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|cccc||cccccccc|} \hline
\neg ... | Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction/Proof by Truth Table | https://proofwiki.org/wiki/Non-Equivalence_as_Conjunction_of_Disjunction_with_Negation_of_Conjunction | https://proofwiki.org/wiki/Non-Equivalence_as_Conjunction_of_Disjunction_with_Negation_of_Conjunction/Proof_by_Truth_Table | [
"Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction",
"Conjunction",
"Disjunction",
"Logical Negation",
"Biconditional"
] | [
"Definition:Logical Not",
"Definition:Biconditional",
"Definition:Exclusive Or"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7208 | Non-Equivalence as Conjunction of Disjunction with Disjunction of Negations | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \lor q} \land \paren {\neg p \lor \neg q}$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {p \lor q} \land \paren {\neg p \lor \neg q} }}
{{Premise |1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\paren {p \lor q} \land \neg \paren {p \land q}|1|Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction}}
{{DeMorgan |3|1|\paren {p \l... | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \lor q} \land \paren {\neg p \lor \neg q}$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {p \lor q} \land \paren {\neg p \lor \neg q} }}
{{Premise |1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\paren {p \lor q} \land \neg \paren {p \land q}|1|[[Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction]]}}
{{DeMorgan |3|1|\paren {... | Non-Equivalence as Conjunction of Disjunction with Disjunction of Negations/Proof 1 | https://proofwiki.org/wiki/Non-Equivalence_as_Conjunction_of_Disjunction_with_Disjunction_of_Negations | https://proofwiki.org/wiki/Non-Equivalence_as_Conjunction_of_Disjunction_with_Disjunction_of_Negations/Proof_1 | [
"Non-Equivalence as Conjunction of Disjunction with Disjunction of Negations",
"Conjunction",
"Disjunction",
"Logical Negation",
"Biconditional"
] | [] | [
"Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction",
"Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction"
] |
proofwiki-7209 | Non-Equivalence as Conjunction of Disjunction with Disjunction of Negations | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \lor q} \land \paren {\neg p \lor \neg q}$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|cccc||ccccccccc|} \hline
\neg & (p & \iff & q) & (p & \lor & q) & \land & (\neg & p & \lor & \neg & q) \\
\hline
\F & \F & \T & \F & \F & \F & \F & \F & \... | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \lor q} \land \paren {\neg p \lor \neg q}$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|cccc||ccccccccc|} \hline
\neg... | Non-Equivalence as Conjunction of Disjunction with Disjunction of Negations/Proof by Truth Table | https://proofwiki.org/wiki/Non-Equivalence_as_Conjunction_of_Disjunction_with_Disjunction_of_Negations | https://proofwiki.org/wiki/Non-Equivalence_as_Conjunction_of_Disjunction_with_Disjunction_of_Negations/Proof_by_Truth_Table | [
"Non-Equivalence as Conjunction of Disjunction with Disjunction of Negations",
"Conjunction",
"Disjunction",
"Logical Negation",
"Biconditional"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7210 | Inversion Mapping is Involution | Let $G$ be a group, and let $\iota: G \to G$ be the inversion mapping.
Then $\iota$ is an involution.
That is:
:$\forall g \in G: \map \iota {\map \iota g} = g$ | Let $g \in G$.
Then:
{{begin-eqn}}
{{eqn | l = \map \iota {\map \iota g}
| r = \paren {g^{-1} }^{-1}
| c = {{Defof|Inversion Mapping}}
}}
{{eqn | r = g
| c = Inverse of Group Inverse
}}
{{end-eqn}}
which establishes the result.
{{qed}}
Category:Inversion Mappings
scu8viz33uzskvkzuh04emh1ycpbec3 | Let $G$ be a [[Definition:Group|group]], and let $\iota: G \to G$ be the [[Definition:Inversion Mapping|inversion mapping]].
Then $\iota$ is an [[Definition:Involution (Mapping)|involution]].
That is:
:$\forall g \in G: \map \iota {\map \iota g} = g$ | Let $g \in G$.
Then:
{{begin-eqn}}
{{eqn | l = \map \iota {\map \iota g}
| r = \paren {g^{-1} }^{-1}
| c = {{Defof|Inversion Mapping}}
}}
{{eqn | r = g
| c = [[Inverse of Group Inverse]]
}}
{{end-eqn}}
which establishes the result.
{{qed}}
[[Category:Inversion Mappings]]
scu8viz33uzskvkzuh04emh1yc... | Inversion Mapping is Involution | https://proofwiki.org/wiki/Inversion_Mapping_is_Involution | https://proofwiki.org/wiki/Inversion_Mapping_is_Involution | [
"Inversion Mappings"
] | [
"Definition:Group",
"Definition:Inversion Mapping",
"Definition:Involution (Mapping)"
] | [
"Inverse of Group Inverse",
"Category:Inversion Mappings"
] |
proofwiki-7211 | Order Isomorphism between Linearly Ordered Spaces is Homeomorphism | Let $\struct {S_1, \le_1, \tau_1}$ and $\struct {S_2, \le_2, \tau_2}$ be linearly ordered spaces.
Let $\phi: S_1 \to S_2$ be an order isomorphism from $\struct {S_1, \le_1}$ to $\struct {S_2, \le_2}$.
Then $\phi$ is a homeomorphism from $\struct {S_1, \tau_1}$ to $\struct {S_2, \tau_2}$. | By the definition of order isomorphism, $\phi$ is a bijection.
Thus to show $\phi$ is a homeomorphism it remains to be shown that:
:$\phi$ is continuous
and:
:$\phi^{-1}$ is continuous.
First it is shown that $\phi^{-1}$ is continuous.
By Order Isomorphism Preserves Initial Segments and its dual, $\phi$ maps open rays ... | Let $\struct {S_1, \le_1, \tau_1}$ and $\struct {S_2, \le_2, \tau_2}$ be [[Definition:Linearly Ordered Space|linearly ordered spaces]].
Let $\phi: S_1 \to S_2$ be an [[Definition:Order Isomorphism|order isomorphism]] from $\struct {S_1, \le_1}$ to $\struct {S_2, \le_2}$.
Then $\phi$ is a [[Definition:Homeomorphism (... | By the definition of [[Definition:Order Isomorphism|order isomorphism]], $\phi$ is a [[Definition:Bijection|bijection]].
Thus to show $\phi$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]] it remains to be shown that:
:$\phi$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]]
an... | Order Isomorphism between Linearly Ordered Spaces is Homeomorphism | https://proofwiki.org/wiki/Order_Isomorphism_between_Linearly_Ordered_Spaces_is_Homeomorphism | https://proofwiki.org/wiki/Order_Isomorphism_between_Linearly_Ordered_Spaces_is_Homeomorphism | [
"Linearly Ordered Spaces"
] | [
"Definition:Linearly Ordered Space",
"Definition:Order Isomorphism",
"Definition:Homeomorphism/Topological Spaces"
] | [
"Definition:Order Isomorphism",
"Definition:Bijection",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Order Isomorphism Preserves Initial... |
proofwiki-7212 | Rule of Association/Conjunction | Conjunction is associative:
=== Formulation 1 ===
{{:Rule of Association/Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Association/Conjunction/Formulation 2}} | {{BeginTableau|p \land \paren {q \land r} \vdash \paren {p \land q} \land r}}
{{Premise|1|p \land \paren {q \land r} }}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q \land r|1|2}}
{{Simplification|4|1|q|3|1}}
{{Simplification|5|1|r|3|2}}
{{Conjunction|6|1|p \land q|2|4}}
{{Conjunction|7|1|\paren {p \land q} \land... | [[Definition:Conjunction|Conjunction]] is [[Definition:Associative Operation|associative]]:
=== [[Rule of Association/Conjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Association/Conjunction/Formulation 1}}
=== [[Rule of Association/Conjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Association/Conjuncti... | {{BeginTableau|p \land \paren {q \land r} \vdash \paren {p \land q} \land r}}
{{Premise|1|p \land \paren {q \land r} }}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q \land r|1|2}}
{{Simplification|4|1|q|3|1}}
{{Simplification|5|1|r|3|2}}
{{Conjunction|6|1|p \land q|2|4}}
{{Conjunction|7|1|\paren {p \land q} \land... | Rule of Association/Conjunction/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Association/Conjunction | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_1/Proof_1 | [
"Rule of Association",
"Conjunction"
] | [
"Definition:Conjunction",
"Definition:Associative Operation",
"Rule of Association/Conjunction/Formulation 1",
"Rule of Association/Conjunction/Formulation 2"
] | [] |
proofwiki-7213 | Rule of Association/Conjunction | Conjunction is associative:
=== Formulation 1 ===
{{:Rule of Association/Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Association/Conjunction/Formulation 2}} | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccc||ccccc|} \hline
p & \land & (q & \land & r) & (p & \land & q) & \land & r \\
\hline
\F & \F & \F & \F & \F & \F & \F & \F & \F & \F \\
\F &... | [[Definition:Conjunction|Conjunction]] is [[Definition:Associative Operation|associative]]:
=== [[Rule of Association/Conjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Association/Conjunction/Formulation 1}}
=== [[Rule of Association/Conjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Association/Conjuncti... | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|ccccc||ccccc|} \hline... | Rule of Association/Conjunction/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Association/Conjunction | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_1/Proof_by_Truth_Table | [
"Rule of Association",
"Conjunction"
] | [
"Definition:Conjunction",
"Definition:Associative Operation",
"Rule of Association/Conjunction/Formulation 1",
"Rule of Association/Conjunction/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7214 | Rule of Association/Conjunction | Conjunction is associative:
=== Formulation 1 ===
{{:Rule of Association/Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Association/Conjunction/Formulation 2}} | === Forward Implication ===
{{:Rule of Association/Conjunction/Formulation 2/Proof 1/Forward Implication}}
=== Reverse Implication ===
{{:Rule of Association/Conjunction/Formulation 2/Proof 1/Reverse Implication}}
{{BeginTableau|\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r} }}
{{T... | [[Definition:Conjunction|Conjunction]] is [[Definition:Associative Operation|associative]]:
=== [[Rule of Association/Conjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Association/Conjunction/Formulation 1}}
=== [[Rule of Association/Conjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Association/Conjuncti... | === [[Rule of Association/Conjunction/Formulation 2/Proof 1/Forward Implication|Forward Implication]] ===
{{:Rule of Association/Conjunction/Formulation 2/Proof 1/Forward Implication}}
=== [[Rule of Association/Conjunction/Formulation 2/Proof 1/Reverse Implication|Reverse Implication]] ===
{{:Rule of Association/Conju... | Rule of Association/Conjunction/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Association/Conjunction | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_2/Proof_1 | [
"Rule of Association",
"Conjunction"
] | [
"Definition:Conjunction",
"Definition:Associative Operation",
"Rule of Association/Conjunction/Formulation 1",
"Rule of Association/Conjunction/Formulation 2"
] | [
"Rule of Association/Conjunction/Formulation 2/Proof 1/Forward Implication",
"Rule of Association/Conjunction/Formulation 2/Proof 1/Reverse Implication",
"Rule of Association/Conjunction/Formulation 2/Proof 1/Forward Implication",
"Rule of Association/Conjunction/Formulation 2/Proof 1/Reverse Implication"
] |
proofwiki-7215 | Rule of Association/Conjunction | Conjunction is associative:
=== Formulation 1 ===
{{:Rule of Association/Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Association/Conjunction/Formulation 2}} | {{BeginTableau|\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r} }}
{{Assumption |1|p \land \paren {q \land r} }}
{{SequentIntro|2|1|\paren {p \land q} \land r|1|Rule of Association: Formulation 1}}
{{Implication |3||\paren {p \land \paren {q \land r} } \implies \paren {\paren {p \la... | [[Definition:Conjunction|Conjunction]] is [[Definition:Associative Operation|associative]]:
=== [[Rule of Association/Conjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Association/Conjunction/Formulation 1}}
=== [[Rule of Association/Conjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Association/Conjuncti... | {{BeginTableau|\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r} }}
{{Assumption |1|p \land \paren {q \land r} }}
{{SequentIntro|2|1|\paren {p \land q} \land r|1|[[Rule of Association/Conjunction/Formulation 1|Rule of Association: Formulation 1]]}}
{{Implication |3||\paren {p \land \... | Rule of Association/Conjunction/Formulation 2/Proof 2 | https://proofwiki.org/wiki/Rule_of_Association/Conjunction | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_2/Proof_2 | [
"Rule of Association",
"Conjunction"
] | [
"Definition:Conjunction",
"Definition:Associative Operation",
"Rule of Association/Conjunction/Formulation 1",
"Rule of Association/Conjunction/Formulation 2"
] | [
"Rule of Association/Conjunction/Formulation 1",
"Rule of Association/Conjunction/Formulation 1"
] |
proofwiki-7216 | Rule of Association/Disjunction | Disjunction is associative:
=== Formulation 1 ===
{{:Rule of Association/Disjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Association/Disjunction/Formulation 2}} | {{BeginTableau|p \lor \paren {q \lor r} \vdash \paren {p \lor q} \lor r}}
{{Premise|1|p \lor \paren {q \lor r} }}
{{Assumption|2|p|By assuming the first main disjunct ...}}
{{Addition|3|2|p \lor q|2|1}}
{{Addition|4|2|\paren {p \lor q} \lor r|3|1|... the conclusion is derived}}
{{Assumption|5|q \lor r|Then assume the s... | [[Definition:Disjunction|Disjunction]] is [[Definition:Associative Operation|associative]]:
=== [[Rule of Association/Disjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Association/Disjunction/Formulation 1}}
=== [[Rule of Association/Disjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Association/Disjuncti... | {{BeginTableau|p \lor \paren {q \lor r} \vdash \paren {p \lor q} \lor r}}
{{Premise|1|p \lor \paren {q \lor r} }}
{{Assumption|2|p|By assuming the first main disjunct ...}}
{{Addition|3|2|p \lor q|2|1}}
{{Addition|4|2|\paren {p \lor q} \lor r|3|1|... the conclusion is derived}}
{{Assumption|5|q \lor r|Then assume the s... | Rule of Association/Disjunction/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Association/Disjunction | https://proofwiki.org/wiki/Rule_of_Association/Disjunction/Formulation_1/Proof_1 | [
"Rule of Association",
"Disjunction"
] | [
"Definition:Disjunction",
"Definition:Associative Operation",
"Rule of Association/Disjunction/Formulation 1",
"Rule of Association/Disjunction/Formulation 2"
] | [] |
proofwiki-7217 | Rule of Association/Disjunction | Disjunction is associative:
=== Formulation 1 ===
{{:Rule of Association/Disjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Association/Disjunction/Formulation 2}} | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccc||ccccc|} \hline
p & \lor & (q & \lor & r) & (p & \lor & q) & \lor & r \\
\hline
\F & \F & \F & \F & \F & \F & \F & \F & \F & \F \\
\F & \T ... | [[Definition:Disjunction|Disjunction]] is [[Definition:Associative Operation|associative]]:
=== [[Rule of Association/Disjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Association/Disjunction/Formulation 1}}
=== [[Rule of Association/Disjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Association/Disjuncti... | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|ccccc||ccccc|} \hline... | Rule of Association/Disjunction/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Association/Disjunction | https://proofwiki.org/wiki/Rule_of_Association/Disjunction/Formulation_1/Proof_by_Truth_Table | [
"Rule of Association",
"Disjunction"
] | [
"Definition:Disjunction",
"Definition:Associative Operation",
"Rule of Association/Disjunction/Formulation 1",
"Rule of Association/Disjunction/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7218 | Rule of Association/Disjunction | Disjunction is associative:
=== Formulation 1 ===
{{:Rule of Association/Disjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Association/Disjunction/Formulation 2}} | {{BeginTableau|\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r} }}
{{Assumption |1|p \lor \paren {q \lor r} }}
{{SequentIntro|2|1|\paren {p \lor q} \lor r|1|Rule of Association: Formulation 1}}
{{Implication |3||\paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor ... | [[Definition:Disjunction|Disjunction]] is [[Definition:Associative Operation|associative]]:
=== [[Rule of Association/Disjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Association/Disjunction/Formulation 1}}
=== [[Rule of Association/Disjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Association/Disjuncti... | {{BeginTableau|\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r} }}
{{Assumption |1|p \lor \paren {q \lor r} }}
{{SequentIntro|2|1|\paren {p \lor q} \lor r|1|[[Rule of Association/Disjunction/Formulation 1|Rule of Association: Formulation 1]]}}
{{Implication |3||\paren {p \lor \paren {q ... | Rule of Association/Disjunction/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Association/Disjunction | https://proofwiki.org/wiki/Rule_of_Association/Disjunction/Formulation_2/Proof_1 | [
"Rule of Association",
"Disjunction"
] | [
"Definition:Disjunction",
"Definition:Associative Operation",
"Rule of Association/Disjunction/Formulation 1",
"Rule of Association/Disjunction/Formulation 2"
] | [
"Rule of Association/Disjunction/Formulation 1",
"Rule of Association/Disjunction/Formulation 1"
] |
proofwiki-7219 | Rule of Association/Disjunction | Disjunction is associative:
=== Formulation 1 ===
{{:Rule of Association/Disjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Association/Disjunction/Formulation 2}} | {{BeginTableau|\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}|Instance 2 of the Hilbert-style systems}}
{{TableauLine
| n = 1
| f = \paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}
| rlnk = Rule of Association/Disjunction/Formulation 2/Forward Implicatio... | [[Definition:Disjunction|Disjunction]] is [[Definition:Associative Operation|associative]]:
=== [[Rule of Association/Disjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Association/Disjunction/Formulation 1}}
=== [[Rule of Association/Disjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Association/Disjuncti... | {{BeginTableau|\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}|[[Definition:Hilbert Proof System/Instance 2|Instance 2 of the Hilbert-style systems]]}}
{{TableauLine
| n = 1
| f = \paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}
| rlnk = Rule of Associati... | Rule of Association/Disjunction/Formulation 2/Proof 2 | https://proofwiki.org/wiki/Rule_of_Association/Disjunction | https://proofwiki.org/wiki/Rule_of_Association/Disjunction/Formulation_2/Proof_2 | [
"Rule of Association",
"Disjunction"
] | [
"Definition:Disjunction",
"Definition:Associative Operation",
"Rule of Association/Disjunction/Formulation 1",
"Rule of Association/Disjunction/Formulation 2"
] | [
"Definition:Hilbert Proof System/Instance 2"
] |
proofwiki-7220 | Rule of Association/Conjunction/Formulation 1 | :$p \land \paren {q \land r} \dashv \vdash \paren {p \land q} \land r$ | {{BeginTableau|p \land \paren {q \land r} \vdash \paren {p \land q} \land r}}
{{Premise|1|p \land \paren {q \land r} }}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q \land r|1|2}}
{{Simplification|4|1|q|3|1}}
{{Simplification|5|1|r|3|2}}
{{Conjunction|6|1|p \land q|2|4}}
{{Conjunction|7|1|\paren {p \land q} \land... | :$p \land \paren {q \land r} \dashv \vdash \paren {p \land q} \land r$ | {{BeginTableau|p \land \paren {q \land r} \vdash \paren {p \land q} \land r}}
{{Premise|1|p \land \paren {q \land r} }}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q \land r|1|2}}
{{Simplification|4|1|q|3|1}}
{{Simplification|5|1|r|3|2}}
{{Conjunction|6|1|p \land q|2|4}}
{{Conjunction|7|1|\paren {p \land q} \land... | Rule of Association/Conjunction/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_1/Proof_1 | [
"Rule of Association"
] | [] | [] |
proofwiki-7221 | Rule of Association/Conjunction/Formulation 1 | :$p \land \paren {q \land r} \dashv \vdash \paren {p \land q} \land r$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccc||ccccc|} \hline
p & \land & (q & \land & r) & (p & \land & q) & \land & r \\
\hline
\F & \F & \F & \F & \F & \F & \F & \F & \F & \F \\
\F &... | :$p \land \paren {q \land r} \dashv \vdash \paren {p \land q} \land r$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|ccccc||ccccc|} \hline... | Rule of Association/Conjunction/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_1/Proof_by_Truth_Table | [
"Rule of Association"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7222 | Rule of Association/Disjunction/Formulation 1 | :$p \lor \paren {q \lor r} \dashv \vdash \paren {p \lor q} \lor r$ | {{BeginTableau|p \lor \paren {q \lor r} \vdash \paren {p \lor q} \lor r}}
{{Premise|1|p \lor \paren {q \lor r} }}
{{Assumption|2|p|By assuming the first main disjunct ...}}
{{Addition|3|2|p \lor q|2|1}}
{{Addition|4|2|\paren {p \lor q} \lor r|3|1|... the conclusion is derived}}
{{Assumption|5|q \lor r|Then assume the s... | :$p \lor \paren {q \lor r} \dashv \vdash \paren {p \lor q} \lor r$ | {{BeginTableau|p \lor \paren {q \lor r} \vdash \paren {p \lor q} \lor r}}
{{Premise|1|p \lor \paren {q \lor r} }}
{{Assumption|2|p|By assuming the first main disjunct ...}}
{{Addition|3|2|p \lor q|2|1}}
{{Addition|4|2|\paren {p \lor q} \lor r|3|1|... the conclusion is derived}}
{{Assumption|5|q \lor r|Then assume the s... | Rule of Association/Disjunction/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Association/Disjunction/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Association/Disjunction/Formulation_1/Proof_1 | [
"Rule of Association"
] | [] | [] |
proofwiki-7223 | Rule of Association/Disjunction/Formulation 1 | :$p \lor \paren {q \lor r} \dashv \vdash \paren {p \lor q} \lor r$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccc||ccccc|} \hline
p & \lor & (q & \lor & r) & (p & \lor & q) & \lor & r \\
\hline
\F & \F & \F & \F & \F & \F & \F & \F & \F & \F \\
\F & \T ... | :$p \lor \paren {q \lor r} \dashv \vdash \paren {p \lor q} \lor r$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|ccccc||ccccc|} \hline... | Rule of Association/Disjunction/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Association/Disjunction/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Association/Disjunction/Formulation_1/Proof_by_Truth_Table | [
"Rule of Association"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7224 | Rule of Association/Conjunction/Formulation 1/Proof 1 | :$p \land \paren {q \land r} \dashv \vdash \paren {p \land q} \land r$ | {{BeginTableau|p \land \paren {q \land r} \vdash \paren {p \land q} \land r}}
{{Premise|1|p \land \paren {q \land r} }}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q \land r|1|2}}
{{Simplification|4|1|q|3|1}}
{{Simplification|5|1|r|3|2}}
{{Conjunction|6|1|p \land q|2|4}}
{{Conjunction|7|1|\paren {p \land q} \land... | :$p \land \paren {q \land r} \dashv \vdash \paren {p \land q} \land r$ | {{BeginTableau|p \land \paren {q \land r} \vdash \paren {p \land q} \land r}}
{{Premise|1|p \land \paren {q \land r} }}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q \land r|1|2}}
{{Simplification|4|1|q|3|1}}
{{Simplification|5|1|r|3|2}}
{{Conjunction|6|1|p \land q|2|4}}
{{Conjunction|7|1|\paren {p \land q} \land... | Rule of Association/Conjunction/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_1/Proof_1 | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_1/Proof_1 | [
"Rule of Association"
] | [] | [] |
proofwiki-7225 | Rule of Association/Conjunction/Formulation 2 | :$\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r}$ | === Forward Implication ===
{{:Rule of Association/Conjunction/Formulation 2/Proof 1/Forward Implication}}
=== Reverse Implication ===
{{:Rule of Association/Conjunction/Formulation 2/Proof 1/Reverse Implication}}
{{BeginTableau|\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r} }}
{{T... | :$\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r}$ | === [[Rule of Association/Conjunction/Formulation 2/Proof 1/Forward Implication|Forward Implication]] ===
{{:Rule of Association/Conjunction/Formulation 2/Proof 1/Forward Implication}}
=== [[Rule of Association/Conjunction/Formulation 2/Proof 1/Reverse Implication|Reverse Implication]] ===
{{:Rule of Association/Conju... | Rule of Association/Conjunction/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_2/Proof_1 | [
"Rule of Association"
] | [] | [
"Rule of Association/Conjunction/Formulation 2/Proof 1/Forward Implication",
"Rule of Association/Conjunction/Formulation 2/Proof 1/Reverse Implication",
"Rule of Association/Conjunction/Formulation 2/Proof 1/Forward Implication",
"Rule of Association/Conjunction/Formulation 2/Proof 1/Reverse Implication"
] |
proofwiki-7226 | Rule of Association/Conjunction/Formulation 2 | :$\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r}$ | {{BeginTableau|\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r} }}
{{Assumption |1|p \land \paren {q \land r} }}
{{SequentIntro|2|1|\paren {p \land q} \land r|1|Rule of Association: Formulation 1}}
{{Implication |3||\paren {p \land \paren {q \land r} } \implies \paren {\paren {p \la... | :$\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r}$ | {{BeginTableau|\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r} }}
{{Assumption |1|p \land \paren {q \land r} }}
{{SequentIntro|2|1|\paren {p \land q} \land r|1|[[Rule of Association/Conjunction/Formulation 1|Rule of Association: Formulation 1]]}}
{{Implication |3||\paren {p \land \... | Rule of Association/Conjunction/Formulation 2/Proof 2 | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_2/Proof_2 | [
"Rule of Association"
] | [] | [
"Rule of Association/Conjunction/Formulation 1",
"Rule of Association/Conjunction/Formulation 1"
] |
proofwiki-7227 | Rule of Association/Disjunction/Formulation 2 | :$\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$ | {{BeginTableau|\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r} }}
{{Assumption |1|p \lor \paren {q \lor r} }}
{{SequentIntro|2|1|\paren {p \lor q} \lor r|1|Rule of Association: Formulation 1}}
{{Implication |3||\paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor ... | :$\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$ | {{BeginTableau|\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r} }}
{{Assumption |1|p \lor \paren {q \lor r} }}
{{SequentIntro|2|1|\paren {p \lor q} \lor r|1|[[Rule of Association/Disjunction/Formulation 1|Rule of Association: Formulation 1]]}}
{{Implication |3||\paren {p \lor \paren {q ... | Rule of Association/Disjunction/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Association/Disjunction/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Association/Disjunction/Formulation_2/Proof_1 | [
"Rule of Association"
] | [] | [
"Rule of Association/Disjunction/Formulation 1",
"Rule of Association/Disjunction/Formulation 1"
] |
proofwiki-7228 | Rule of Association/Disjunction/Formulation 2 | :$\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$ | {{BeginTableau|\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}|Instance 2 of the Hilbert-style systems}}
{{TableauLine
| n = 1
| f = \paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}
| rlnk = Rule of Association/Disjunction/Formulation 2/Forward Implicatio... | :$\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$ | {{BeginTableau|\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}|[[Definition:Hilbert Proof System/Instance 2|Instance 2 of the Hilbert-style systems]]}}
{{TableauLine
| n = 1
| f = \paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}
| rlnk = Rule of Associati... | Rule of Association/Disjunction/Formulation 2/Proof 2 | https://proofwiki.org/wiki/Rule_of_Association/Disjunction/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Association/Disjunction/Formulation_2/Proof_2 | [
"Rule of Association"
] | [] | [
"Definition:Hilbert Proof System/Instance 2"
] |
proofwiki-7229 | Principle of Commutation/Formulation 1/Forward Implication | :$p \implies \paren {q \implies r} \vdash q \implies \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \implies r} \vdash q \implies \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \implies r} }}
{{Assumption|2|q}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q \implies r|1|3}}
{{ModusPonens|5|1, 2, 3|r|2|4}}
{{Implication|6|1, 2|p \implies r|3|5}}
{{Implication|7|1|q \implies \p... | :$p \implies \paren {q \implies r} \vdash q \implies \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \implies r} \vdash q \implies \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \implies r} }}
{{Assumption|2|q}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q \implies r|1|3}}
{{ModusPonens|5|1, 2, 3|r|2|4}}
{{Implication|6|1, 2|p \implies r|3|5}}
{{Implication|7|1|q \implies \p... | Principle of Commutation/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_1/Forward_Implication/Proof | [
"Principle of Commutation"
] | [] | [] |
proofwiki-7230 | Principle of Commutation/Formulation 1 | :$p \implies \paren {q \implies r} \dashv \vdash q \implies \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \implies r} \vdash q \implies \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \implies r} }}
{{Assumption|2|q}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q \implies r|1|3}}
{{ModusPonens|5|1, 2, 3|r|2|4}}
{{Implication|6|1, 2|p \implies r|3|5}}
{{Implication|7|1|q \implies \p... | :$p \implies \paren {q \implies r} \dashv \vdash q \implies \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \implies r} \vdash q \implies \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \implies r} }}
{{Assumption|2|q}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q \implies r|1|3}}
{{ModusPonens|5|1, 2, 3|r|2|4}}
{{Implication|6|1, 2|p \implies r|3|5}}
{{Implication|7|1|q \implies \p... | Principle of Commutation/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_1 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_1/Forward_Implication/Proof | [
"Principle of Commutation"
] | [] | [] |
proofwiki-7231 | Principle of Commutation/Formulation 1 | :$p \implies \paren {q \implies r} \dashv \vdash q \implies \paren {p \implies r}$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccc||ccccc|} \hline
p & \implies & (q & \implies & r) & q & \implies & (p & \implies & r) \\
\hline
\F & \T & \F & \T & \F &... | :$p \implies \paren {q \implies r} \dashv \vdash q \implies \paren {p \implies r}$ | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|cc... | Principle of Commutation/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_1 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_1/Proof_by_Truth_Table | [
"Principle of Commutation"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7232 | Principle of Commutation/Formulation 1 | :$p \implies \paren {q \implies r} \dashv \vdash q \implies \paren {p \implies r}$ | {{BeginTableau|q \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r} }}
{{Premise|1|q \implies \paren {p \implies r} }}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{ModusPonens|4|1, 3|p \implies r|1|3}}
{{ModusPonens|5|1, 2, 3|r|2|4}}
{{Implication|6|1, 2|q \implies r|3|5}}
{{Implication|7|1|p \implies \p... | :$p \implies \paren {q \implies r} \dashv \vdash q \implies \paren {p \implies r}$ | {{BeginTableau|q \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r} }}
{{Premise|1|q \implies \paren {p \implies r} }}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{ModusPonens|4|1, 3|p \implies r|1|3}}
{{ModusPonens|5|1, 2, 3|r|2|4}}
{{Implication|6|1, 2|q \implies r|3|5}}
{{Implication|7|1|p \implies \p... | Principle of Commutation/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_1 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_1/Reverse_Implication/Proof | [
"Principle of Commutation"
] | [] | [] |
proofwiki-7233 | Principle of Commutation/Formulation 1/Reverse Implication | :$q \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r}$ | {{BeginTableau|q \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r} }}
{{Premise|1|q \implies \paren {p \implies r} }}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{ModusPonens|4|1, 3|p \implies r|1|3}}
{{ModusPonens|5|1, 2, 3|r|2|4}}
{{Implication|6|1, 2|q \implies r|3|5}}
{{Implication|7|1|p \implies \p... | :$q \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r}$ | {{BeginTableau|q \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r} }}
{{Premise|1|q \implies \paren {p \implies r} }}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{ModusPonens|4|1, 3|p \implies r|1|3}}
{{ModusPonens|5|1, 2, 3|r|2|4}}
{{Implication|6|1, 2|q \implies r|3|5}}
{{Implication|7|1|p \implies \p... | Principle of Commutation/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_1/Reverse_Implication/Proof | [
"Principle of Commutation"
] | [] | [] |
proofwiki-7234 | Principle of Commutation/Formulation 2 | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} } }}
{{TheoremIntro|1|\paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} }|Principle of Commutation: Formulation 2: Forward Implication}}
{{TheoremIntro|2|\paren {q \i... | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} } }}
{{TheoremIntro|1|\paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} }|[[Principle of Commutation/Formulation 2/Forward Implication/Proof 1|Principle of Commutatio... | Principle of Commutation/Formulation 2 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2 | [
"Principle of Commutation"
] | [] | [
"Principle of Commutation/Formulation 2/Forward Implication/Proof 1",
"Principle of Commutation/Formulation 2/Reverse Implication/Proof"
] |
proofwiki-7235 | Principle of Commutation/Formulation 2 | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{SequentIntro|2|1|q \implies \paren {p \implies r}|1|Principle of Commutation: Formulation 1: Forward Implication}}
{{Implication|3||\paren {p \impl... | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{SequentIntro|2|1|q \implies \paren {p \implies r}|1|[[Principle of Commutation/Formulation 1/Forward Implication|Principle of Commutation: Formulat... | Principle of Commutation/Formulation 2/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2/Forward_Implication/Proof_1 | [
"Principle of Commutation"
] | [] | [
"Principle of Commutation/Formulation 1/Forward Implication"
] |
proofwiki-7236 | Principle of Commutation/Formulation 2 | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{Assumption|2|q}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q \implies r|1|3}}
{{ModusPonens|5|1, 2, 3|r|2|4}}
{{Implication|6|1, 2|p \implies r|3|5}}... | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{Assumption|2|q}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q \implies r|1|3}}
{{ModusPonens|5|1, 2, 3|r|2|4}}
{{Implication|6|1, 2|p \implies r|3|5}}... | Principle of Commutation/Formulation 2/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2/Forward_Implication/Proof_2 | [
"Principle of Commutation"
] | [] | [] |
proofwiki-7237 | Principle of Commutation/Formulation 2 | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$ | Using a tableau proof for instance 1 of a Hilbert proof system:
{{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } |nohead = 1}}
{{Assumption|1|p}}
{{Assumption|2|p \implies \paren {q \implies r} }}
{{ModusPonens|3|1, 2|q \implies r|1|2}}
{{Assumption|4|... | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$ | Using a [[Definition:Tableau Proof (Formal Systems)|tableau proof]] for [[Definition:Hilbert Proof System/Instance 1|instance 1 of a Hilbert proof system]]:
{{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } |nohead = 1}}
{{Assumption|1|p}}
{{Assumption... | Principle of Commutation/Formulation 2/Forward Implication/Proof 3 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2/Forward_Implication/Proof_3 | [
"Principle of Commutation"
] | [] | [
"Definition:Tableau Proof (Natural Deduction)",
"Definition:Hilbert Proof System/Instance 1"
] |
proofwiki-7238 | Principle of Commutation/Formulation 2 | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption |1|q \implies \paren {p \implies r} }}
{{SequentIntro |2|1|p \implies \paren {q \implies r}|1|Principle of Commutation: Formulation 1: Reverse Implication}}
{{Implication |3| |\paren {... | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption |1|q \implies \paren {p \implies r} }}
{{SequentIntro |2|1|p \implies \paren {q \implies r}|1|[[Principle of Commutation/Formulation 1/Reverse Implication|Principle of Commutation: Form... | Principle of Commutation/Formulation 2/Reverse Implication/Proof | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2/Reverse_Implication/Proof | [
"Principle of Commutation"
] | [] | [
"Principle of Commutation/Formulation 1/Reverse Implication"
] |
proofwiki-7239 | Principle of Commutation/Formulation 2/Forward Implication | :$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{SequentIntro|2|1|q \implies \paren {p \implies r}|1|Principle of Commutation: Formulation 1: Forward Implication}}
{{Implication|3||\paren {p \impl... | :$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{SequentIntro|2|1|q \implies \paren {p \implies r}|1|[[Principle of Commutation/Formulation 1/Forward Implication|Principle of Commutation: Formulat... | Principle of Commutation/Formulation 2/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2/Forward_Implication/Proof_1 | [
"Principle of Commutation"
] | [] | [
"Principle of Commutation/Formulation 1/Forward Implication"
] |
proofwiki-7240 | Principle of Commutation/Formulation 2/Forward Implication | :$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{Assumption|2|q}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q \implies r|1|3}}
{{ModusPonens|5|1, 2, 3|r|2|4}}
{{Implication|6|1, 2|p \implies r|3|5}}... | :$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{Assumption|2|q}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q \implies r|1|3}}
{{ModusPonens|5|1, 2, 3|r|2|4}}
{{Implication|6|1, 2|p \implies r|3|5}}... | Principle of Commutation/Formulation 2/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2/Forward_Implication/Proof_2 | [
"Principle of Commutation"
] | [] | [] |
proofwiki-7241 | Principle of Commutation/Formulation 2/Forward Implication | :$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} }$ | Using a tableau proof for instance 1 of a Hilbert proof system:
{{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } |nohead = 1}}
{{Assumption|1|p}}
{{Assumption|2|p \implies \paren {q \implies r} }}
{{ModusPonens|3|1, 2|q \implies r|1|2}}
{{Assumption|4|... | :$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} }$ | Using a [[Definition:Tableau Proof (Formal Systems)|tableau proof]] for [[Definition:Hilbert Proof System/Instance 1|instance 1 of a Hilbert proof system]]:
{{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } |nohead = 1}}
{{Assumption|1|p}}
{{Assumption... | Principle of Commutation/Formulation 2/Forward Implication/Proof 3 | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2/Forward_Implication/Proof_3 | [
"Principle of Commutation"
] | [] | [
"Definition:Tableau Proof (Natural Deduction)",
"Definition:Hilbert Proof System/Instance 1"
] |
proofwiki-7242 | Principle of Commutation/Formulation 2/Reverse Implication | :$\vdash \paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption |1|q \implies \paren {p \implies r} }}
{{SequentIntro |2|1|p \implies \paren {q \implies r}|1|Principle of Commutation: Formulation 1: Reverse Implication}}
{{Implication |3| |\paren {... | :$\vdash \paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption |1|q \implies \paren {p \implies r} }}
{{SequentIntro |2|1|p \implies \paren {q \implies r}|1|[[Principle of Commutation/Formulation 1/Reverse Implication|Principle of Commutation: Form... | Principle of Commutation/Formulation 2/Reverse Implication/Proof | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Principle_of_Commutation/Formulation_2/Reverse_Implication/Proof | [
"Principle of Commutation"
] | [] | [
"Principle of Commutation/Formulation 1/Reverse Implication"
] |
proofwiki-7243 | Linearly Ordered Space is Connected iff Linear Continuum | Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Then $S$ is a connected space {{iff}} it is a linear continuum. | {{explain|It is not clear how the premises of the subproofs (e.g. "$T$ is disconnected and densely ordered" directly relate to the premises of the main theorem, e.g.: "$S$ is a connected space..."}} | Let $T = \struct {S, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Then $S$ is a [[Definition:Connected Topological Space|connected space]] {{iff}} it is a [[Definition:Linear Continuum|linear continuum]]. | {{explain|It is not clear how the premises of the subproofs (e.g. "$T$ is [[Definition:Disconnected Space|disconnected]] and [[Definition:Densely Ordered|densely ordered]]" directly relate to the premises of the main theorem, e.g.: "$S$ is a [[Definition:Connected Topological Space|connected space]]..."}} | Linearly Ordered Space is Connected iff Linear Continuum | https://proofwiki.org/wiki/Linearly_Ordered_Space_is_Connected_iff_Linear_Continuum | https://proofwiki.org/wiki/Linearly_Ordered_Space_is_Connected_iff_Linear_Continuum | [
"Linearly Ordered Spaces",
"Examples of Connected Topological Spaces",
"Examples of Linear Continua"
] | [
"Definition:Linearly Ordered Space",
"Definition:Connected Topological Space",
"Definition:Linear Continuum"
] | [
"Definition:Disconnected (Topology)/Topological Space",
"Definition:Densely Ordered",
"Definition:Connected Topological Space",
"Definition:Disconnected (Topology)/Topological Space",
"Definition:Densely Ordered",
"Definition:Densely Ordered",
"Definition:Densely Ordered",
"Definition:Densely Ordered"... |
proofwiki-7244 | Rule of Transposition/Variant 1/Formulation 2/Forward Implication | :$\vdash \paren {p \implies \neg q} \implies \paren {q \implies \neg p}$ | {{BeginTableau|\vdash \paren {p \implies \neg q} \implies \paren {q \implies \neg p} }}
{{Assumption|1|p \implies \neg q}}
{{Assumption|2|q}}
{{DoubleNegIntro|3|2|\neg \neg q|2}}
{{ModusTollens|4|1, 2|\neg p|1|3}}
{{Implication|5|1|q \implies \neg p|2|4}}
{{Implication|6||\paren {p \implies \neg q} \implies \paren {q \... | :$\vdash \paren {p \implies \neg q} \implies \paren {q \implies \neg p}$ | {{BeginTableau|\vdash \paren {p \implies \neg q} \implies \paren {q \implies \neg p} }}
{{Assumption|1|p \implies \neg q}}
{{Assumption|2|q}}
{{DoubleNegIntro|3|2|\neg \neg q|2}}
{{ModusTollens|4|1, 2|\neg p|1|3}}
{{Implication|5|1|q \implies \neg p|2|4}}
{{Implication|6||\paren {p \implies \neg q} \implies \paren {q \... | Rule of Transposition/Variant 1/Formulation 2/Forward Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_2/Forward_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7245 | Boundary of Polygon is Jordan Curve | Let $P$ be a polygon embedded in $\R^2$.
Then there exists a Jordan curve $\gamma: \closedint 0 1 \to \R^2$ such that the image of $\gamma$ is equal to the boundary $\partial P$ of $P$. | The polygon $P$ has $n$ sides, where $n \in \N$.
Denote the vertices of $P$ as $A_1, \ldots, A_n$ and its sides as $S_1, \ldots, S_n$, such that each vertex $A_i$ has adjacent sides $S_{i - 1}$ and $S_i$.
We use the conventions that $S_0 = S_n$, and $A_{n + 1} = A_1$.
As each side $S_i$ is a line segment joining $A_i$ ... | Let $P$ be a [[Definition:Polygon|polygon]] embedded in $\R^2$.
Then there exists a [[Definition:Jordan Curve|Jordan curve]] $\gamma: \closedint 0 1 \to \R^2$ such that the [[Definition:Image of Mapping|image]] of $\gamma$ is equal to the [[Definition:Boundary (Geometry)|boundary]] $\partial P$ of $P$. | The [[Definition:Polygon|polygon]] $P$ has $n$ [[Definition:Side of Polygon|sides]], where $n \in \N$.
Denote the [[Definition:Vertex of Polygon|vertices]] of $P$ as $A_1, \ldots, A_n$ and its sides as $S_1, \ldots, S_n$, such that each [[Definition:Vertex of Polygon|vertex]] $A_i$ has [[Definition:Adjacent Side to Ve... | Boundary of Polygon is Jordan Curve | https://proofwiki.org/wiki/Boundary_of_Polygon_is_Jordan_Curve | https://proofwiki.org/wiki/Boundary_of_Polygon_is_Jordan_Curve | [
"Jordan Curves"
] | [
"Definition:Polygon",
"Definition:Jordan Curve",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Boundary (Geometry)"
] | [
"Definition:Polygon",
"Definition:Polygon/Side",
"Definition:Polygon/Vertex",
"Definition:Polygon/Vertex",
"Definition:Polygon/Adjacent/Side to Vertex",
"Definition:Convex Set (Vector Space)/Line Segment",
"Definition:Path (Topology)",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Pol... |
proofwiki-7246 | Rule of Explosion/Variant 1 | :$\vdash p \implies \paren {\neg p \implies q}$ | {{BeginTableau|\vdash p \implies \paren {\neg p \implies q} }}
{{Assumption|1|p}}
{{Assumption|2|\neg p}}
{{NonContradiction|3|1, 2|1|2}}
{{Explosion|4|1, 2|q|3}}
{{Implication|5|1|\neg p \implies q|2|4}}
{{Implication|6||p \implies \paren {\neg p \implies q}|1|5}}
{{EndTableau|qed}} | :$\vdash p \implies \paren {\neg p \implies q}$ | {{BeginTableau|\vdash p \implies \paren {\neg p \implies q} }}
{{Assumption|1|p}}
{{Assumption|2|\neg p}}
{{NonContradiction|3|1, 2|1|2}}
{{Explosion|4|1, 2|q|3}}
{{Implication|5|1|\neg p \implies q|2|4}}
{{Implication|6||p \implies \paren {\neg p \implies q}|1|5}}
{{EndTableau|qed}} | Rule of Explosion/Variant 1 | https://proofwiki.org/wiki/Rule_of_Explosion/Variant_1 | https://proofwiki.org/wiki/Rule_of_Explosion/Variant_1 | [
"Rule of Explosion"
] | [] | [] |
proofwiki-7247 | Negation of Conditional implies Antecedent | :$\vdash \neg \paren {p \implies q} \implies p$ | {{BeginTableau|\vdash \neg \paren {p \implies q} \implies p}}
{{Assumption|1|\neg \paren {p \implies q} }}
{{Assumption|2|\neg \paren {p \land \neg q} }}
{{SequentIntro|3|2|p \implies q|2|Conditional is Equivalent to Negation of Conjunction with Negative}}
{{NonContradiction|4|1, 2|3|1}}
{{Reductio|5|1|p \land \neg q|2... | :$\vdash \neg \paren {p \implies q} \implies p$ | {{BeginTableau|\vdash \neg \paren {p \implies q} \implies p}}
{{Assumption|1|\neg \paren {p \implies q} }}
{{Assumption|2|\neg \paren {p \land \neg q} }}
{{SequentIntro|3|2|p \implies q|2|[[Conditional is Equivalent to Negation of Conjunction with Negative]]}}
{{NonContradiction|4|1, 2|3|1}}
{{Reductio|5|1|p \land \neg... | Negation of Conditional implies Antecedent/Proof 1 | https://proofwiki.org/wiki/Negation_of_Conditional_implies_Antecedent | https://proofwiki.org/wiki/Negation_of_Conditional_implies_Antecedent/Proof_1 | [
"Negation of Conditional implies Antecedent",
"Conditional",
"Logical Negation"
] | [] | [
"Conditional is Equivalent to Negation of Conjunction with Negative"
] |
proofwiki-7248 | Negation of Conditional implies Antecedent | :$\vdash \neg \paren {p \implies q} \implies p$ | {{BeginTableau|\vdash \neg \paren {p \implies q} \implies p}}
{{Assumption|1|\neg \paren {p \implies q} }}
{{SequentIntro|2|1|p \land \neg q|1|Conjunction with Negative is Equivalent to Negation of Conditional}}
{{Simplification|3|1|p|2|1}}
{{Implication|4||\neg \paren {p \implies q} \implies p|1|3}}
{{EndTableau|qed}} | :$\vdash \neg \paren {p \implies q} \implies p$ | {{BeginTableau|\vdash \neg \paren {p \implies q} \implies p}}
{{Assumption|1|\neg \paren {p \implies q} }}
{{SequentIntro|2|1|p \land \neg q|1|[[Conjunction with Negative is Equivalent to Negation of Conditional]]}}
{{Simplification|3|1|p|2|1}}
{{Implication|4||\neg \paren {p \implies q} \implies p|1|3}}
{{EndTableau|q... | Negation of Conditional implies Antecedent/Proof 2 | https://proofwiki.org/wiki/Negation_of_Conditional_implies_Antecedent | https://proofwiki.org/wiki/Negation_of_Conditional_implies_Antecedent/Proof_2 | [
"Negation of Conditional implies Antecedent",
"Conditional",
"Logical Negation"
] | [] | [
"Conjunction with Negative is Equivalent to Negation of Conditional"
] |
proofwiki-7249 | Negation of Conditional implies Negation of Consequent | :$\vdash \neg \paren {p \implies q} \implies \neg q$ | {{BeginTableau|\vdash \neg \paren {p \implies q} \implies \neg q}}
{{Assumption|1|\neg \paren {p \implies q} }}
{{Assumption|2|\neg \paren {p \land \neg q} }}
{{SequentIntro|3|2|p \implies q|2|Conditional is Equivalent to Negation of Conjunction with Negative}}
{{NonContradiction|4|1, 2|3|1}}
{{Reductio|5|1|p \land \ne... | :$\vdash \neg \paren {p \implies q} \implies \neg q$ | {{BeginTableau|\vdash \neg \paren {p \implies q} \implies \neg q}}
{{Assumption|1|\neg \paren {p \implies q} }}
{{Assumption|2|\neg \paren {p \land \neg q} }}
{{SequentIntro|3|2|p \implies q|2|[[Conditional is Equivalent to Negation of Conjunction with Negative]]}}
{{NonContradiction|4|1, 2|3|1}}
{{Reductio|5|1|p \land... | Negation of Conditional implies Negation of Consequent/Proof 1 | https://proofwiki.org/wiki/Negation_of_Conditional_implies_Negation_of_Consequent | https://proofwiki.org/wiki/Negation_of_Conditional_implies_Negation_of_Consequent/Proof_1 | [
"Negation of Conditional implies Negation of Consequent",
"Conditional",
"Logical Negation"
] | [] | [
"Conditional is Equivalent to Negation of Conjunction with Negative"
] |
proofwiki-7250 | Negation of Conditional implies Negation of Consequent | :$\vdash \neg \paren {p \implies q} \implies \neg q$ | {{BeginTableau|\vdash \neg \paren {p \implies q} \implies \neg q}}
{{Assumption|1|\neg \paren {p \implies q} }}
{{SequentIntro|2|1|p \land \neg q|1|Conjunction with Negative is Equivalent to Negation of Conditional}}
{{Simplification|3|1|\neg q|2|2}}
{{Implication|4||\neg \paren {p \implies q} \implies \neg q|1|3}}
{{E... | :$\vdash \neg \paren {p \implies q} \implies \neg q$ | {{BeginTableau|\vdash \neg \paren {p \implies q} \implies \neg q}}
{{Assumption|1|\neg \paren {p \implies q} }}
{{SequentIntro|2|1|p \land \neg q|1|[[Conjunction with Negative is Equivalent to Negation of Conditional]]}}
{{Simplification|3|1|\neg q|2|2}}
{{Implication|4||\neg \paren {p \implies q} \implies \neg q|1|3}}... | Negation of Conditional implies Negation of Consequent/Proof 2 | https://proofwiki.org/wiki/Negation_of_Conditional_implies_Negation_of_Consequent | https://proofwiki.org/wiki/Negation_of_Conditional_implies_Negation_of_Consequent/Proof_2 | [
"Negation of Conditional implies Negation of Consequent",
"Conditional",
"Logical Negation"
] | [] | [
"Conjunction with Negative is Equivalent to Negation of Conditional"
] |
proofwiki-7251 | Matrix Multiplication Interpretation of Relation Composition | Let $A$, $B$ and $C$ be finite non-empty sets that are initial segments of $\N_{\ne 0}$.
Let $\RR \subseteq B \times A$ and $\SS \subseteq C \times B$ be relations.
Let $\mathbf R$ and $\mathbf S$ be matrices which we define as follows:
:$\sqbrk r_{i j} = \begin{cases}
\T & : \tuple {i, j} \in \RR \\
\F & : \tuple {i, ... | === Sufficient Condition ===
Suppose for some $i, j$:
:$\sqbrk {r s}_{i j} = \T$
Then by definition of $\lor$ there must exist some $k$ for which:
:$\sqbrk r_{i k} \land \sqbrk s_{k j} = \T$
which by our definition implies:
:$\tuple {i, k} \in \RR$
:$\tuple {k, j} \in \SS$
Then by definition of a composite relation:
:$... | Let $A$, $B$ and $C$ be [[Definition:Finite Set|finite]] [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|sets]] that are [[Definition:Initial Segment|initial segments]] of $\N_{\ne 0}$.
Let $\RR \subseteq B \times A$ and $\SS \subseteq C \times B$ be [[Definition:Relation|relations]].
Let $\mathbf R$ and $\ma... | === Sufficient Condition ===
Suppose for some $i, j$:
:$\sqbrk {r s}_{i j} = \T$
Then by definition of $\lor$ there must exist some $k$ for which:
:$\sqbrk r_{i k} \land \sqbrk s_{k j} = \T$
which by our definition implies:
:$\tuple {i, k} \in \RR$
:$\tuple {k, j} \in \SS$
Then by definition of a [[Definition:... | Matrix Multiplication Interpretation of Relation Composition | https://proofwiki.org/wiki/Matrix_Multiplication_Interpretation_of_Relation_Composition | https://proofwiki.org/wiki/Matrix_Multiplication_Interpretation_of_Relation_Composition | [
"Relation Theory"
] | [
"Definition:Finite Set",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Initial Segment",
"Definition:Relation",
"Definition:Matrix",
"Definition:Matrix Product (Conventional)",
"Definition:Composition of Relations",
"Definition:Ring (Abstract Algebra)",
"Definition:Matrix Product (Conv... | [
"Definition:Composition of Relations"
] |
proofwiki-7252 | Space is Compact iff Every Cover from Basis has Finite Subcover | Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB \subseteq \tau$ be a basis of $\tau$.
Then $T = \struct {S, \tau}$ is '''compact''' {{iff}}:
:from every cover of $S$ by elements of $\BB$, a finite subcover of $S$ can be selected. | === Sufficient Condition ===
Let every open cover for $S$ have a finite subcover.
Then every cover of $S$ by elements of $\BB$ is an open cover for $S$.
So:
:from every cover of $S$ by elements of $\BB$, a finite subcover of $S$ can be selected.
{{qed|lemma}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\BB \subseteq \tau$ be a [[Definition:Basis (Topology)|basis]] of $\tau$.
Then $T = \struct {S, \tau}$ is '''[[Definition:Compact Topological Space|compact]]''' {{iff}}:
:from every [[Definition:Cover of Set|cover]] of $S$ by [... | === Sufficient Condition ===
Let every [[Definition:Open Cover|open cover]] for $S$ have a [[Definition:Finite Subcover|finite subcover]].
Then every [[Definition:Cover of Set|cover]] of $S$ by [[Definition:Element|elements]] of $\BB$ is an [[Definition:Open Cover|open cover]] for $S$.
So:
:from every [[Definition:C... | Space is Compact iff Every Cover from Basis has Finite Subcover | https://proofwiki.org/wiki/Space_is_Compact_iff_Every_Cover_from_Basis_has_Finite_Subcover | https://proofwiki.org/wiki/Space_is_Compact_iff_Every_Cover_from_Basis_has_Finite_Subcover | [
"Compact Topological Spaces",
"Subcovers",
"Topological Bases"
] | [
"Definition:Topological Space",
"Definition:Basis (Topology)",
"Definition:Compact Topological Space",
"Definition:Cover of Set",
"Definition:Element",
"Definition:Subcover/Finite"
] | [
"Definition:Open Cover",
"Definition:Subcover/Finite",
"Definition:Cover of Set",
"Definition:Element",
"Definition:Open Cover",
"Definition:Cover of Set",
"Definition:Element",
"Definition:Subcover/Finite",
"Definition:Cover of Set",
"Definition:Element",
"Definition:Subcover/Finite",
"Defini... |
proofwiki-7253 | Set of Subsets is Cover iff Set of Complements is Free | Let $S$ be a set.
Let $\CC$ be a set of sets.
Then $\CC$ is a cover for $S$ {{iff}} $\set {\relcomp S X: X \in \CC}$ is free. | Let $S$ be a set.
Let us recall the definition of free:
{{:Definition:Free Set of Sets}} | Let $S$ be a [[Definition:Set|set]].
Let $\CC$ be a [[Definition:Set of Sets|set of sets]].
Then $\CC$ is a [[Definition:Cover of Set|cover for $S$]] {{iff}} $\set {\relcomp S X: X \in \CC}$ is [[Definition:Free Set of Sets|free]]. | Let $S$ be a [[Definition:Set|set]].
Let us recall the definition of [[Definition:Free Set of Sets|free]]:
{{:Definition:Free Set of Sets}} | Set of Subsets is Cover iff Set of Complements is Free | https://proofwiki.org/wiki/Set_of_Subsets_is_Cover_iff_Set_of_Complements_is_Free | https://proofwiki.org/wiki/Set_of_Subsets_is_Cover_iff_Set_of_Complements_is_Free | [
"Set Theory"
] | [
"Definition:Set",
"Definition:Set of Sets",
"Definition:Cover of Set",
"Definition:Free Set of Sets"
] | [
"Definition:Set",
"Definition:Free Set of Sets",
"Definition:Free Set of Sets",
"Definition:Free Set of Sets",
"Definition:Free Set of Sets"
] |
proofwiki-7254 | Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 2/Reverse Implication | :$\vdash \paren {\neg \paren {p \land \neg q} } \implies \paren {p \implies q}$ | {{BeginTableau|\vdash \paren {\neg \paren {p \land \neg q} } \implies \paren {p \implies q} }}
{{Assumption|1|\neg \paren {p \land \neg q} }}
{{SequentIntro|2|1|p \implies q|1|Conditional is Equivalent to Negation of Conjunction with Negative: Formulation 1}}
{{Implication|3||\paren {\neg \paren {p \land \neg q} } \imp... | :$\vdash \paren {\neg \paren {p \land \neg q} } \implies \paren {p \implies q}$ | {{BeginTableau|\vdash \paren {\neg \paren {p \land \neg q} } \implies \paren {p \implies q} }}
{{Assumption|1|\neg \paren {p \land \neg q} }}
{{SequentIntro|2|1|p \implies q|1|[[Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Reverse Implication|Conditional is Equivalent to Negation of ... | Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 2/Reverse Implication | https://proofwiki.org/wiki/Conditional_is_Equivalent_to_Negation_of_Conjunction_with_Negative/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Conditional_is_Equivalent_to_Negation_of_Conjunction_with_Negative/Formulation_2/Reverse_Implication | [
"Conditional is Equivalent to Negation of Conjunction with Negative"
] | [] | [
"Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Reverse Implication",
"Category:Conditional is Equivalent to Negation of Conjunction with Negative"
] |
proofwiki-7255 | Biconditional Elimination | The '''rule of biconditional elimination''' is a valid argument in types of logic dealing with conditionals $\implies$ and biconditionals $\iff$.
This includes classical propositional logic and predicate logic, and in particular natural deduction.
=== Proof Rule ===
{{:Biconditional Elimination/Proof Rule}}
=== Sequent... | === Form 1 ===
{{:Biconditional Elimination/Sequent Form/Proof 1/Form 1}}
=== Form 2 ===
{{:Biconditional Elimination/Sequent Form/Proof 1/Form 2}} | The '''rule of [[Biconditional Elimination|biconditional elimination]]''' is a [[Definition:Valid Argument|valid argument]] in types of [[Definition:Logic|logic]] dealing with [[Definition:Conditional|conditionals]] $\implies$ and [[Definition:Biconditional|biconditionals]] $\iff$.
This includes [[Definition:Classical... | === [[Biconditional Elimination/Sequent Form/Proof 1/Form 1|Form 1]] ===
{{:Biconditional Elimination/Sequent Form/Proof 1/Form 1}}
=== [[Biconditional Elimination/Sequent Form/Proof 1/Form 2|Form 2]] ===
{{:Biconditional Elimination/Sequent Form/Proof 1/Form 2}} | Biconditional Elimination/Sequent Form/Proof 1 | https://proofwiki.org/wiki/Biconditional_Elimination | https://proofwiki.org/wiki/Biconditional_Elimination/Sequent_Form/Proof_1 | [
"Biconditional Elimination",
"Biconditional",
"Conditional"
] | [
"Biconditional Elimination",
"Definition:Valid Argument",
"Definition:Logic",
"Definition:Conditional",
"Definition:Biconditional",
"Definition:Classical Propositional Logic",
"Definition:Predicate Logic",
"Definition:Natural Deduction",
"Biconditional Elimination/Proof Rule",
"Biconditional Elimi... | [
"Biconditional Elimination/Sequent Form/Proof 1/Form 1",
"Biconditional Elimination/Sequent Form/Proof 1/Form 2"
] |
proofwiki-7256 | Biconditional Elimination | The '''rule of biconditional elimination''' is a valid argument in types of logic dealing with conditionals $\implies$ and biconditionals $\iff$.
This includes classical propositional logic and predicate logic, and in particular natural deduction.
=== Proof Rule ===
{{:Biconditional Elimination/Proof Rule}}
=== Sequent... | We apply the Method of Truth Tables.
$\begin{array}{|ccc||ccc|ccc|} \hline
p & \iff & q & p & \implies & q & q & \implies & p \\
\hline
\F & \T & \F & \F & \T & \F & \F & \T & \F \\
\F & \F & \T & \F & \T & \T & \T & \F & \F \\
\T & \F & \F & \T & \F & \F & \F & \T & \T \\
\T & \T & \T & \T & \F & \T & \T & \T & \T \\
... | The '''rule of [[Biconditional Elimination|biconditional elimination]]''' is a [[Definition:Valid Argument|valid argument]] in types of [[Definition:Logic|logic]] dealing with [[Definition:Conditional|conditionals]] $\implies$ and [[Definition:Biconditional|biconditionals]] $\iff$.
This includes [[Definition:Classical... | We apply the [[Method of Truth Tables]].
$\begin{array}{|ccc||ccc|ccc|} \hline
p & \iff & q & p & \implies & q & q & \implies & p \\
\hline
\F & \T & \F & \F & \T & \F & \F & \T & \F \\
\F & \F & \T & \F & \T & \T & \T & \F & \F \\
\T & \F & \F & \T & \F & \F & \F & \T & \T \\
\T & \T & \T & \T & \F & \T & \T & \T & \... | Biconditional Elimination/Sequent Form/Proof by Truth Table | https://proofwiki.org/wiki/Biconditional_Elimination | https://proofwiki.org/wiki/Biconditional_Elimination/Sequent_Form/Proof_by_Truth_Table | [
"Biconditional Elimination",
"Biconditional",
"Conditional"
] | [
"Biconditional Elimination",
"Definition:Valid Argument",
"Definition:Logic",
"Definition:Conditional",
"Definition:Biconditional",
"Definition:Classical Propositional Logic",
"Definition:Predicate Logic",
"Definition:Natural Deduction",
"Biconditional Elimination/Proof Rule",
"Biconditional Elimi... | [
"Method of Truth Tables",
"Definition:True"
] |
proofwiki-7257 | Biconditional Introduction/Sequent Form | {{begin-eqn}}
{{eqn | l = p
| o = \implies
| r = q
}}
{{eqn | l = q
| o = \implies
| r = p
}}
{{eqn | ll= \vdash
| l = p
| o = \iff
| r = q
}}
{{end-eqn}} | {{BeginTableau|p \implies q, q \implies p \vdash p \iff q}}
{{Premise|1|p \implies q}}
{{Premise|2|q \implies p}}
{{BiconditionalIntro|3|1, 2|p \iff q|1|2}}
{{EndTableau}}
{{Qed}} | {{begin-eqn}}
{{eqn | l = p
| o = \implies
| r = q
}}
{{eqn | l = q
| o = \implies
| r = p
}}
{{eqn | ll= \vdash
| l = p
| o = \iff
| r = q
}}
{{end-eqn}} | {{BeginTableau|p \implies q, q \implies p \vdash p \iff q}}
{{Premise|1|p \implies q}}
{{Premise|2|q \implies p}}
{{BiconditionalIntro|3|1, 2|p \iff q|1|2}}
{{EndTableau}}
{{Qed}} | Biconditional Introduction/Sequent Form/Proof 1 | https://proofwiki.org/wiki/Biconditional_Introduction/Sequent_Form | https://proofwiki.org/wiki/Biconditional_Introduction/Sequent_Form/Proof_1 | [
"Biconditional Introduction"
] | [] | [] |
proofwiki-7258 | Biconditional Introduction/Sequent Form | {{begin-eqn}}
{{eqn | l = p
| o = \implies
| r = q
}}
{{eqn | l = q
| o = \implies
| r = p
}}
{{eqn | ll= \vdash
| l = p
| o = \iff
| r = q
}}
{{end-eqn}} | We apply the Method of Truth Tables.
:<nowiki>$\begin {array} {|ccccccc||ccc|} \hline
(p & \implies & q) & \land & (q & \implies & p) & p & \iff & q\\
\hline
\F & \T & \F & \T & \F & \T & \F & \F & \T & \F \\
\F & \T & \T & \F & \T & \F & \F & \F & \F & \T \\
\T & \F & \F & \F & \F & \T & \T & \T & \F & \F \\
\T & \T &... | {{begin-eqn}}
{{eqn | l = p
| o = \implies
| r = q
}}
{{eqn | l = q
| o = \implies
| r = p
}}
{{eqn | ll= \vdash
| l = p
| o = \iff
| r = q
}}
{{end-eqn}} | We apply the [[Method of Truth Tables]].
:<nowiki>$\begin {array} {|ccccccc||ccc|} \hline
(p & \implies & q) & \land & (q & \implies & p) & p & \iff & q\\
\hline
\F & \T & \F & \T & \F & \T & \F & \F & \T & \F \\
\F & \T & \T & \F & \T & \F & \F & \F & \F & \T \\
\T & \F & \F & \F & \F & \T & \T & \T & \F & \F \\
\T &... | Biconditional Introduction/Sequent Form/Proof by Truth Table | https://proofwiki.org/wiki/Biconditional_Introduction/Sequent_Form | https://proofwiki.org/wiki/Biconditional_Introduction/Sequent_Form/Proof_by_Truth_Table | [
"Biconditional Introduction"
] | [] | [
"Method of Truth Tables",
"Definition:True"
] |
proofwiki-7259 | Biconditional Elimination/Sequent Form/Proof 1/Form 1 | :$p \iff q \vdash p \implies q$ | {{BeginTableau|p \iff q \vdash p \implies q}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{EndTableau}}
{{Qed}} | :$p \iff q \vdash p \implies q$ | {{BeginTableau|p \iff q \vdash p \implies q}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{EndTableau}}
{{Qed}} | Biconditional Elimination/Sequent Form/Proof 1/Form 1 | https://proofwiki.org/wiki/Biconditional_Elimination/Sequent_Form/Proof_1/Form_1 | https://proofwiki.org/wiki/Biconditional_Elimination/Sequent_Form/Proof_1/Form_1 | [
"Biconditional Elimination"
] | [] | [] |
proofwiki-7260 | Biconditional Elimination/Sequent Form/Proof 1/Form 2 | :$p \iff q \vdash q \implies p$ | {{BeginTableau|p \iff q \vdash q \implies p}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|q \implies p|1|2}}
{{EndTableau}}
{{Qed}} | :$p \iff q \vdash q \implies p$ | {{BeginTableau|p \iff q \vdash q \implies p}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|q \implies p|1|2}}
{{EndTableau}}
{{Qed}} | Biconditional Elimination/Sequent Form/Proof 1/Form 2 | https://proofwiki.org/wiki/Biconditional_Elimination/Sequent_Form/Proof_1/Form_2 | https://proofwiki.org/wiki/Biconditional_Elimination/Sequent_Form/Proof_1/Form_2 | [
"Biconditional Elimination"
] | [] | [] |
proofwiki-7261 | Biconditional is Commutative/Formulation 1/Proof 1 | : $p \iff q \dashv \vdash q \iff p$ | {{BeginTableau|p \iff q \vdash q \iff p}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{BiconditionalElimination|3|1|q \implies p|1|2}}
{{BiconditionalIntro|4|1|q \iff p|3|2}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|q \iff p \vdash p \iff q}}
{{Premise|1|q \iff p}}
{{BiconditionalElimin... | : $p \iff q \dashv \vdash q \iff p$ | {{BeginTableau|p \iff q \vdash q \iff p}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{BiconditionalElimination|3|1|q \implies p|1|2}}
{{BiconditionalIntro|4|1|q \iff p|3|2}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|q \iff p \vdash p \iff q}}
{{Premise|1|q \iff p}}
{{BiconditionalElim... | Biconditional is Commutative/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Biconditional_is_Commutative/Formulation_1/Proof_1 | https://proofwiki.org/wiki/Biconditional_is_Commutative/Formulation_1/Proof_1 | [
"Biconditional is Commutative"
] | [] | [] |
proofwiki-7262 | Rule of Material Equivalence/Formulation 1 | :$p \iff q \dashv \vdash \paren {p \implies q} \land \paren {q \implies p}$ | {{BeginTableau|p \iff q \vdash \paren {p \implies q} \land \paren {q \implies p} }}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{BiconditionalElimination|3|1|q \implies p|1|2}}
{{Conjunction|4|1|\paren {p \implies q} \land \paren {q \implies p}|2|3}}
{{EndTableau}}
{{BeginTableau|\paren {p... | :$p \iff q \dashv \vdash \paren {p \implies q} \land \paren {q \implies p}$ | {{BeginTableau|p \iff q \vdash \paren {p \implies q} \land \paren {q \implies p} }}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{BiconditionalElimination|3|1|q \implies p|1|2}}
{{Conjunction|4|1|\paren {p \implies q} \land \paren {q \implies p}|2|3}}
{{EndTableau}}
{{BeginTableau|\paren ... | Rule of Material Equivalence/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_1/Proof_1 | [
"Rule of Material Equivalence"
] | [] | [] |
proofwiki-7263 | Rule of Material Equivalence/Formulation 1 | :$p \iff q \dashv \vdash \paren {p \implies q} \land \paren {q \implies p}$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|ccc|ccccccc|} \hline
p & \iff & q & (p & \implies & q) & \land & (q & \implies & p) \\
\hline
\F & \T & \F & \F & \T & \F & \T & \F & \T & \F \\
... | :$p \iff q \dashv \vdash \paren {p \implies q} \land \paren {q \implies p}$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|ccc|ccccccc|} \hline
... | Rule of Material Equivalence/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_1/Proof_by_Truth_Table | [
"Rule of Material Equivalence"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7264 | Equivalences are Interderivable/Forward Implication | : $\left ({p \dashv \vdash q}\right) \vdash \left ({p \iff q}\right)$ | {{BeginTableau|\left({p \dashv \vdash q}\right) \vdash p \iff q}}
{{Premise|1|p \dashv \vdash q}}
{{SequentIntro|2|1|\left ({p \vdash q}\right) \land \left ({q \vdash p}\right)|1
|Definition of Interderivable}}
{{Assumption|3|p}}
{{Simplification|4|1, 3|p \vdash q|2|1}}
{{Implication|5|1|p \implies q|3|4}}
{{Assumpti... | : $\left ({p \dashv \vdash q}\right) \vdash \left ({p \iff q}\right)$ | {{BeginTableau|\left({p \dashv \vdash q}\right) \vdash p \iff q}}
{{Premise|1|p \dashv \vdash q}}
{{SequentIntro|2|1|\left ({p \vdash q}\right) \land \left ({q \vdash p}\right)|1
|Definition of [[Definition:Interderivable|Interderivable]]}}
{{Assumption|3|p}}
{{Simplification|4|1, 3|p \vdash q|2|1}}
{{Implication|5|1... | Equivalences are Interderivable/Forward Implication | https://proofwiki.org/wiki/Equivalences_are_Interderivable/Forward_Implication | https://proofwiki.org/wiki/Equivalences_are_Interderivable/Forward_Implication | [
"Equivalences are Interderivable"
] | [] | [
"Definition:Logical Equivalence",
"Category:Equivalences are Interderivable"
] |
proofwiki-7265 | Equivalences are Interderivable/Reverse Implication | : $\left ({p \iff q}\right) \vdash \left ({p \dashv \vdash q}\right)$ | {{BeginTableau|p \iff q \vdash \left({p \vdash q}\right)}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|3|1|p \implies q|1|1}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q|2|3}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|p \iff q \vdash \left({q \vdash p}\right)}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|3|... | : $\left ({p \iff q}\right) \vdash \left ({p \dashv \vdash q}\right)$ | {{BeginTableau|p \iff q \vdash \left({p \vdash q}\right)}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|3|1|p \implies q|1|1}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q|2|3}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|p \iff q \vdash \left({q \vdash p}\right)}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|... | Equivalences are Interderivable/Reverse Implication | https://proofwiki.org/wiki/Equivalences_are_Interderivable/Reverse_Implication | https://proofwiki.org/wiki/Equivalences_are_Interderivable/Reverse_Implication | [
"Equivalences are Interderivable"
] | [] | [
"Category:Equivalences are Interderivable"
] |
proofwiki-7266 | Compact Subspace of Linearly Ordered Space/Reverse Implication | Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq X$ be a non-empty subset of $X$.
Let the following hold:
:$(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
:$(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.
Then $Y$ is a co... | Let $\tau'$ be the $\tau$-relative subspace topology on $Y$.
Let $\preceq'$ be the restriction of $\preceq$ to $Y$.
=== Lemma ===
{{:Compact Subspace of Linearly Ordered Space/Lemma 2}}{{qed|lemma}}
The premises immediately show that $\struct {Y, \preceq'}$ is a complete lattice.
By Complete Linearly Ordered Space is C... | Let $\struct {X, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Let $Y \subseteq X$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $X$.
Let the following hold:
:$(1): \quad$ For every [[Definition:Non-Empty Set|non-empty]] $S \subseteq Y$, $S$ has a [[... | Let $\tau'$ be the $\tau$-relative [[Definition:Subspace Topology|subspace topology]] on $Y$.
Let $\preceq'$ be the [[Definition:Restriction of Ordering|restriction]] of $\preceq$ to $Y$.
=== [[Compact Subspace of Linearly Ordered Space/Lemma 2|Lemma]] ===
{{:Compact Subspace of Linearly Ordered Space/Lemma 2}}{{qed... | Compact Subspace of Linearly Ordered Space/Reverse Implication/Proof 1 | https://proofwiki.org/wiki/Compact_Subspace_of_Linearly_Ordered_Space/Reverse_Implication | https://proofwiki.org/wiki/Compact_Subspace_of_Linearly_Ordered_Space/Reverse_Implication/Proof_1 | [
"Compact Subspace of Linearly Ordered Space"
] | [
"Definition:Linearly Ordered Space",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Non-Empty Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Non-Empty Set",
"Definition:Compact Topological Space/Subspace"
] | [
"Definition:Topological Subspace",
"Definition:Restriction of Ordering",
"Compact Subspace of Linearly Ordered Space/Lemma 2",
"Definition:Complete Lattice",
"Complete Linearly Ordered Space is Compact",
"Definition:Compact Topological Space/Subspace"
] |
proofwiki-7267 | Compact Subspace of Linearly Ordered Space/Reverse Implication | Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq X$ be a non-empty subset of $X$.
Let the following hold:
:$(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
:$(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.
Then $Y$ is a co... | Let $\FF$ be an ultrafilter on $Y$.
For $S \in \FF$, let $\map f S = \inf S$.
Let $p = \sup \map f \FF$.
{{improve|perhaps directly use $p :{{=}} \ds \sup_{S \mathop \in \FF} \inf S$?}}
Then $\FF$ converges to $p$:
=== Upward rays ===
Let $a \in X$ with $a \prec p$.
Since $\FF$ is an ultrafilter, either $Y \cap {\uparr... | Let $\struct {X, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Let $Y \subseteq X$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $X$.
Let the following hold:
:$(1): \quad$ For every [[Definition:Non-Empty Set|non-empty]] $S \subseteq Y$, $S$ has a [[... | Let $\FF$ be an [[Definition:Ultrafilter on Set|ultrafilter]] on $Y$.
For $S \in \FF$, let $\map f S = \inf S$.
Let $p = \sup \map f \FF$.
{{improve|perhaps directly use $p :{{=}} \ds \sup_{S \mathop \in \FF} \inf S$?}}
Then $\FF$ converges to $p$:
=== Upward rays ===
Let $a \in X$ with $a \prec p$.
Since $\FF$ ... | Compact Subspace of Linearly Ordered Space/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Compact_Subspace_of_Linearly_Ordered_Space/Reverse_Implication | https://proofwiki.org/wiki/Compact_Subspace_of_Linearly_Ordered_Space/Reverse_Implication/Proof_2 | [
"Compact Subspace of Linearly Ordered Space"
] | [
"Definition:Linearly Ordered Space",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Non-Empty Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Non-Empty Set",
"Definition:Compact Topological Space/Subspace"
] | [
"Definition:Ultrafilter on Set",
"Definition:Ultrafilter on Set",
"Definition:Filter on Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Contradiction",
"Definition:Lower Bound of Set",
"Definition:Infimum of Set",
"Extended Transitivity",
"Definition:Supremum of Set",
... |
proofwiki-7268 | Modus Ponendo Tollens/Sequent Form/Case 1 | :$\neg \paren {p \land q}, p \vdash \neg q$ | {{BeginTableau|\neg \paren {p \land q}, p \vdash \neg q}}
{{Premise|1|\neg \paren {p \land q} }}
{{Premise|2|p}}
{{Assumption|3|q}}
{{Conjunction|4|2, 3|p \land q|2|3}}
{{NonContradiction|5|1, 2, 3|4|1}}
{{Contradiction|6|1, 2|\neg q|3|5}}
{{EndTableau|qed}} | :$\neg \paren {p \land q}, p \vdash \neg q$ | {{BeginTableau|\neg \paren {p \land q}, p \vdash \neg q}}
{{Premise|1|\neg \paren {p \land q} }}
{{Premise|2|p}}
{{Assumption|3|q}}
{{Conjunction|4|2, 3|p \land q|2|3}}
{{NonContradiction|5|1, 2, 3|4|1}}
{{Contradiction|6|1, 2|\neg q|3|5}}
{{EndTableau|qed}} | Modus Ponendo Tollens/Sequent Form/Case 1 | https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Sequent_Form/Case_1 | https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Sequent_Form/Case_1 | [
"Modus Ponendo Tollens"
] | [] | [] |
proofwiki-7269 | Modus Ponendo Tollens/Sequent Form/Case 2 | :$\neg \left({p \land q}\right), q \vdash \neg p$ | {{BeginTableau|\neg \left({p \land q}\right), q \vdash \neg p}}
{{Premise|1|\neg \left({p \land q}\right)}}
{{Premise|2|q}}
{{Assumption|3|p}}
{{Conjunction|4|2, 3|p \land q|3|2}}
{{NonContradiction|5|1, 2, 3|4|1}}
{{Contradiction|6|1, 2|\neg p|3|5}}
{{EndTableau}}
{{qed}}
Category:Modus Ponendo Tollens
mjeey35ngrumpe2... | :$\neg \left({p \land q}\right), q \vdash \neg p$ | {{BeginTableau|\neg \left({p \land q}\right), q \vdash \neg p}}
{{Premise|1|\neg \left({p \land q}\right)}}
{{Premise|2|q}}
{{Assumption|3|p}}
{{Conjunction|4|2, 3|p \land q|3|2}}
{{NonContradiction|5|1, 2, 3|4|1}}
{{Contradiction|6|1, 2|\neg p|3|5}}
{{EndTableau}}
{{qed}}
[[Category:Modus Ponendo Tollens]]
mjeey35ngr... | Modus Ponendo Tollens/Sequent Form/Case 2 | https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Sequent_Form/Case_2 | https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Sequent_Form/Case_2 | [
"Modus Ponendo Tollens"
] | [] | [
"Category:Modus Ponendo Tollens"
] |
proofwiki-7270 | Union of Subsets is Subset/Subset of Power Set | Let $S$ and $T$ be sets.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S$ be a subset of $\powerset S$.
Then:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$ | Let $\mathbb S \subseteq \powerset S$.
Suppose that $\forall X \in \mathbb S: X \subseteq T$.
Consider any $\ds x \in \bigcup \mathbb S$.
By definition of set union, it follows that:
:$\exists X \in \mathbb S: x \in X$
But as $X \subseteq T$ it follows that $x \in T$.
Thus it follows that:
:$\ds \bigcup \mathbb S \subs... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\mathbb S$ be a [[Definition:Subset|subset]] of $\powerset S$.
Then:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$ | Let $\mathbb S \subseteq \powerset S$.
Suppose that $\forall X \in \mathbb S: X \subseteq T$.
Consider any $\ds x \in \bigcup \mathbb S$.
By definition of [[Definition:Set Union|set union]], it follows that:
:$\exists X \in \mathbb S: x \in X$
But as $X \subseteq T$ it follows that $x \in T$.
Thus it follows that:... | Union of Subsets is Subset/Subset of Power Set | https://proofwiki.org/wiki/Union_of_Subsets_is_Subset/Subset_of_Power_Set | https://proofwiki.org/wiki/Union_of_Subsets_is_Subset/Subset_of_Power_Set | [
"Set Union",
"Subsets"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Subset"
] | [
"Definition:Set Union",
"Category:Set Union",
"Category:Subsets"
] |
proofwiki-7271 | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2/Forward Implication | :$\vdash \paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} }$ | {{BeginTableau|\vdash \paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} } }}
{{Assumption|1|p \land \neg q}}
{{Assumption|2|p \implies q| Assume the opposite of what is to be proved ...}}
{{Simplification|3|1|p|1|1}}
{{Simplification|4|1|\neg q|1|2}}
{{ModusPonens|5|1, 2|q|2|3}}
{{NonContradiction|6|1... | :$\vdash \paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} }$ | {{BeginTableau|\vdash \paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} } }}
{{Assumption|1|p \land \neg q}}
{{Assumption|2|p \implies q| Assume the opposite of what is to be proved ...}}
{{Simplification|3|1|p|1|1}}
{{Simplification|4|1|\neg q|1|2}}
{{ModusPonens|5|1, 2|q|2|3}}
{{NonContradiction|6|1... | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2/Forward_Implication/Proof_1 | [
"Conjunction with Negative is Equivalent to Negation of Conditional"
] | [] | [] |
proofwiki-7272 | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2/Reverse Implication | :$\vdash \paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q}$ | {{BeginTableau|\vdash \paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q} }}
{{Assumption|1|\neg \paren {p \implies q} }}
{{Assumption|2|\neg \paren {p \land \neg q} }}
{{SequentIntro|3|2|p \implies q|2|Conditional is Equivalent to Negation of Conjunction with Negative}}
{{NonContradiction|4|1, 2|3|1}... | :$\vdash \paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q}$ | {{BeginTableau|\vdash \paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q} }}
{{Assumption|1|\neg \paren {p \implies q} }}
{{Assumption|2|\neg \paren {p \land \neg q} }}
{{SequentIntro|3|2|p \implies q|2|[[Conditional is Equivalent to Negation of Conjunction with Negative]]}}
{{NonContradiction|4|1, 2|... | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2/Reverse Implication/Proof 1 | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2/Reverse_Implication/Proof_1 | [
"Conjunction with Negative is Equivalent to Negation of Conditional"
] | [] | [
"Conditional is Equivalent to Negation of Conjunction with Negative"
] |
proofwiki-7273 | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2/Reverse Implication | :$\vdash \paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q}$ | {{BeginTableau|\vdash \paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q} }}
{{Assumption|1|\neg \paren {p \implies q} }}
{{SequentIntro|2|1|p \land \neg q|1|Conjunction with Negative is Equivalent to Negation of Conditional: Formulation 1: Reverse Implication}}
{{Implication|3||\paren {\neg \paren {p... | :$\vdash \paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q}$ | {{BeginTableau|\vdash \paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q} }}
{{Assumption|1|\neg \paren {p \implies q} }}
{{SequentIntro|2|1|p \land \neg q|1|[[Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1/Reverse Implication|Conjunction with Negative is Equivalent t... | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2/Reverse_Implication/Proof_2 | [
"Conjunction with Negative is Equivalent to Negation of Conditional"
] | [] | [
"Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1/Reverse Implication"
] |
proofwiki-7274 | Convex Set is Star Convex Set | Let $V$ be a vector space over $\R$ or $\C$.
Let $A \subseteq V$ be a non-empty convex set.
Then $A$ is a star convex set, and every point in $A$ is a star center. | Let $a \in A$.
Note that there is at least one point in $A$, as $A$ is non-empty.
If $x \in A$, then there is a line segment joining $a$ and $x$.
By definition, it follows that $A$ is star convex, and $a$ is a star center.
{{qed}}
Category:Convex Sets (Vector Spaces)
bnnv4ei5bq7etdgcg2soj8upxict53j | Let $V$ be a [[Definition:Vector Space|vector space]] over $\R$ or $\C$.
Let $A \subseteq V$ be a [[Definition:Empty Set|non-empty]] [[Definition:Convex Set (Vector Space)|convex set]].
Then $A$ is a [[Definition:Star Convex Set|star convex set]], and every point in $A$ is a [[Definition:Star Convex Set|star center]... | Let $a \in A$.
Note that there is at least one point in $A$, as $A$ is [[Definition:Empty Set|non-empty]].
If $x \in A$, then there is a [[Definition:Straight Line Segment (Vector Space)|line segment]] joining $a$ and $x$.
By definition, it follows that $A$ is [[Definition:Star Convex Set|star convex]], and $a$ is a... | Convex Set is Star Convex Set | https://proofwiki.org/wiki/Convex_Set_is_Star_Convex_Set | https://proofwiki.org/wiki/Convex_Set_is_Star_Convex_Set | [
"Convex Sets (Vector Spaces)"
] | [
"Definition:Vector Space",
"Definition:Empty Set",
"Definition:Convex Set (Vector Space)",
"Definition:Star Convex Set",
"Definition:Star Convex Set"
] | [
"Definition:Empty Set",
"Definition:Convex Set (Vector Space)/Line Segment",
"Definition:Star Convex Set",
"Definition:Star Convex Set",
"Category:Convex Sets (Vector Spaces)"
] |
proofwiki-7275 | Negation implies Negation of Conjunction/Case 1 | :$\neg p \implies \neg \paren {p \land q}$ | {{BeginTableau|\neg p \implies \neg \paren {p \land q} }}
{{Assumption|1|\neg p}}
{{Assumption|2|p \land q}}
{{Simplification|3|2|p|2|1}}
{{NonContradiction|4|1, 2|3|1}}
{{Contradiction|5|1|\neg \paren {p \land q}|2|4}}
{{Implication|6||\neg p \implies \neg \paren {p \land q}|1|5}}
{{EndTableau|qed}} | :$\neg p \implies \neg \paren {p \land q}$ | {{BeginTableau|\neg p \implies \neg \paren {p \land q} }}
{{Assumption|1|\neg p}}
{{Assumption|2|p \land q}}
{{Simplification|3|2|p|2|1}}
{{NonContradiction|4|1, 2|3|1}}
{{Contradiction|5|1|\neg \paren {p \land q}|2|4}}
{{Implication|6||\neg p \implies \neg \paren {p \land q}|1|5}}
{{EndTableau|qed}} | Negation implies Negation of Conjunction/Case 1 | https://proofwiki.org/wiki/Negation_implies_Negation_of_Conjunction/Case_1 | https://proofwiki.org/wiki/Negation_implies_Negation_of_Conjunction/Case_1 | [
"Negation implies Negation of Conjunction"
] | [] | [] |
proofwiki-7276 | Negation implies Negation of Conjunction/Case 2 | :$\neg q \implies \neg \paren {p \land q}$ | {{BeginTableau|\neg q \implies \neg \paren {p \land q} }}
{{Assumption|1|\neg q}}
{{Assumption|2|p \land q}}
{{Simplification|3|2|q|2|2}}
{{NonContradiction|4|1, 2|3|1}}
{{Contradiction|5|1|\neg \paren {p \land q}|2|4}}
{{Implication|6||\neg q \implies \neg \paren {p \land q}|1|5}}
{{EndTableau|qed}} | :$\neg q \implies \neg \paren {p \land q}$ | {{BeginTableau|\neg q \implies \neg \paren {p \land q} }}
{{Assumption|1|\neg q}}
{{Assumption|2|p \land q}}
{{Simplification|3|2|q|2|2}}
{{NonContradiction|4|1, 2|3|1}}
{{Contradiction|5|1|\neg \paren {p \land q}|2|4}}
{{Implication|6||\neg q \implies \neg \paren {p \land q}|1|5}}
{{EndTableau|qed}} | Negation implies Negation of Conjunction/Case 2/Proof | https://proofwiki.org/wiki/Negation_implies_Negation_of_Conjunction/Case_2 | https://proofwiki.org/wiki/Negation_implies_Negation_of_Conjunction/Case_2/Proof | [
"Negation implies Negation of Conjunction"
] | [] | [] |
proofwiki-7277 | Neighborhood Sub-Basis Criterion for Filter Convergence | Let $\struct {S, \tau}$ be a topological space.
Let $\FF$ be a filter on $S$.
Let $p \in S$.
Then $\FF$ converges to $p$ {{iff}} $\FF$ contains as a subset a neighborhood sub-basis at $p$. | === Sufficient Condition ===
Let $\FF$ converges to $p$.
Then it contains ''every'' neighborhood of $p$.
The set of neighborhoods of $p$ is trivially a neighborhood sub-basis at $p$.
{{qed|lemma}} | Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\FF$ be a [[Definition:Filter on Set|filter]] on $S$.
Let $p \in S$.
Then $\FF$ [[Definition:Convergent Filter|converges]] to $p$ {{iff}} $\FF$ contains as a [[Definition:Subset|subset]] a [[Definition:Neighborhood Sub-Basis|neigh... | === Sufficient Condition ===
Let $\FF$ [[Definition:Convergent Filter|converges]] to $p$.
Then it contains ''every'' [[Definition:Neighborhood of Point|neighborhood]] of $p$.
The set of [[Definition:Neighborhood of Point|neighborhoods]] of $p$ is trivially a [[Definition:Neighborhood Sub-Basis|neighborhood sub-basis... | Neighborhood Sub-Basis Criterion for Filter Convergence | https://proofwiki.org/wiki/Neighborhood_Sub-Basis_Criterion_for_Filter_Convergence | https://proofwiki.org/wiki/Neighborhood_Sub-Basis_Criterion_for_Filter_Convergence | [
"Filter Theory",
"Topological Bases"
] | [
"Definition:Topological Space",
"Definition:Filter on Set",
"Definition:Convergent Filter",
"Definition:Subset",
"Definition:Neighborhood Sub-Basis"
] | [
"Definition:Convergent Filter",
"Definition:Neighborhood (Topology)/Point",
"Definition:Neighborhood (Topology)/Point",
"Definition:Neighborhood Sub-Basis",
"Definition:Neighborhood Sub-Basis",
"Definition:Neighborhood (Topology)/Point",
"Definition:Neighborhood Sub-Basis",
"Definition:Neighborhood (T... |
proofwiki-7278 | Rule of Material Implication/Formulation 2/Forward Implication | : $\vdash \left({p \implies q}\right) \implies \left({\neg p \lor q}\right)$ | {{BeginTableau|\left({p \implies q}\right) \implies \left({\neg p \lor q}\right)}}
{{Assumption|1|p \implies q}}
{{SequentIntro|2|1|\neg p \lor q|1|Rule of Material Implication: Formulation 1}}
{{Implication|3||\left({p \implies q}\right) \implies \left({\neg p \lor q}\right)|1|2}}
{{EndTableau}}
{{qed}}
{{LEM|Rule of ... | : $\vdash \left({p \implies q}\right) \implies \left({\neg p \lor q}\right)$ | {{BeginTableau|\left({p \implies q}\right) \implies \left({\neg p \lor q}\right)}}
{{Assumption|1|p \implies q}}
{{SequentIntro|2|1|\neg p \lor q|1|[[Rule of Material Implication/Formulation 1/Forward Implication|Rule of Material Implication: Formulation 1]]}}
{{Implication|3||\left({p \implies q}\right) \implies \left... | Rule of Material Implication/Formulation 2/Forward Implication | https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_2/Forward_Implication | [
"Rule of Material Implication"
] | [] | [
"Rule of Material Implication/Formulation 1/Forward Implication"
] |
proofwiki-7279 | Rule of Material Implication/Formulation 2/Reverse Implication | :$\vdash \paren {\neg p \lor q} \implies \paren {p \implies q}$ | {{BeginTableau|\paren {\neg p \lor q} \implies \paren {p \implies q} }}
{{Assumption|1|\neg p \lor q}}
{{SequentIntro|2|1|p \implies q|1|Rule of Material Implication: Formulation 1}}
{{Implication|3||\paren {\neg p \lor q} \implies \paren {p \implies q}|1|2}}
{{EndTableau}}
{{qed}} | :$\vdash \paren {\neg p \lor q} \implies \paren {p \implies q}$ | {{BeginTableau|\paren {\neg p \lor q} \implies \paren {p \implies q} }}
{{Assumption|1|\neg p \lor q}}
{{SequentIntro|2|1|p \implies q|1|[[Rule of Material Implication/Formulation 1/Reverse Implication|Rule of Material Implication: Formulation 1]]}}
{{Implication|3||\paren {\neg p \lor q} \implies \paren {p \implies q}... | Rule of Material Implication/Formulation 2/Reverse Implication | https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_2/Reverse_Implication | [
"Rule of Material Implication"
] | [] | [
"Rule of Material Implication/Formulation 1/Reverse Implication"
] |
proofwiki-7280 | Compact Subspace of Linearly Ordered Space | Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq X$ be a non-empty subset of $X$.
Then $Y$ is a compact subspace of $\struct {X, \tau}$ {{iff}} both of the following hold:
:$(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
:$(2): \quad$ For every n... | === Forward Implication ===
Let $S$ be a non-empty subset of $Y$.
By Compact Subspace of Linearly Ordered Space: Lemma 1, $S$ has a supremum $k$ in $Y$.
{{explain|The lemma shows that the structure in question is a complete lattice. Take that further step to assert the existence of the supremum.}}
{{explain|supremum WR... | Let $\struct {X, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Let $Y \subseteq X$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $X$.
Then $Y$ is a [[Definition:Compact Subspace|compact subspace]] of $\struct {X, \tau}$ {{iff}} both of the following ... | === Forward Implication ===
Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $Y$.
By [[Compact Subspace of Linearly Ordered Space/Lemma 1|Compact Subspace of Linearly Ordered Space: Lemma 1]], $S$ has a [[Definition:Supremum of Set|supremum]] $k$ in $Y$.
{{explain|The lemma shows t... | Compact Subspace of Linearly Ordered Space | https://proofwiki.org/wiki/Compact_Subspace_of_Linearly_Ordered_Space | https://proofwiki.org/wiki/Compact_Subspace_of_Linearly_Ordered_Space | [
"Compact Subspace of Linearly Ordered Space",
"Linearly Ordered Spaces",
"Compact Topological Spaces"
] | [
"Definition:Linearly Ordered Space",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Compact Topological Space/Subspace",
"Definition:Non-Empty Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Non-Empty Set"
] | [
"Definition:Non-Empty Set",
"Definition:Subset",
"Compact Subspace of Linearly Ordered Space/Lemma 1",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Lower Closure/Element",
"Definition:Upper Closure/Element",
"Definition:Open Cover",
"Definition:Subcover/Finite",
"Defi... |
proofwiki-7281 | Compact Subspace of Linearly Ordered Space | Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq X$ be a non-empty subset of $X$.
Then $Y$ is a compact subspace of $\struct {X, \tau}$ {{iff}} both of the following hold:
:$(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
:$(2): \quad$ For every n... | Let $\tau'$ be the $\tau$-relative subspace topology on $Y$.
Let $\preceq'$ be the restriction of $\preceq$ to $Y$.
=== Lemma ===
{{:Compact Subspace of Linearly Ordered Space/Lemma 2}}{{qed|lemma}}
The premises immediately show that $\struct {Y, \preceq'}$ is a complete lattice.
By Complete Linearly Ordered Space is C... | Let $\struct {X, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Let $Y \subseteq X$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $X$.
Then $Y$ is a [[Definition:Compact Subspace|compact subspace]] of $\struct {X, \tau}$ {{iff}} both of the following ... | Let $\tau'$ be the $\tau$-relative [[Definition:Subspace Topology|subspace topology]] on $Y$.
Let $\preceq'$ be the [[Definition:Restriction of Ordering|restriction]] of $\preceq$ to $Y$.
=== [[Compact Subspace of Linearly Ordered Space/Lemma 2|Lemma]] ===
{{:Compact Subspace of Linearly Ordered Space/Lemma 2}}{{qed... | Compact Subspace of Linearly Ordered Space/Reverse Implication/Proof 1 | https://proofwiki.org/wiki/Compact_Subspace_of_Linearly_Ordered_Space | https://proofwiki.org/wiki/Compact_Subspace_of_Linearly_Ordered_Space/Reverse_Implication/Proof_1 | [
"Compact Subspace of Linearly Ordered Space",
"Linearly Ordered Spaces",
"Compact Topological Spaces"
] | [
"Definition:Linearly Ordered Space",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Compact Topological Space/Subspace",
"Definition:Non-Empty Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Non-Empty Set"
] | [
"Definition:Topological Subspace",
"Definition:Restriction of Ordering",
"Compact Subspace of Linearly Ordered Space/Lemma 2",
"Definition:Complete Lattice",
"Complete Linearly Ordered Space is Compact",
"Definition:Compact Topological Space/Subspace"
] |
proofwiki-7282 | Compact Subspace of Linearly Ordered Space | Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq X$ be a non-empty subset of $X$.
Then $Y$ is a compact subspace of $\struct {X, \tau}$ {{iff}} both of the following hold:
:$(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
:$(2): \quad$ For every n... | Let $\FF$ be an ultrafilter on $Y$.
For $S \in \FF$, let $\map f S = \inf S$.
Let $p = \sup \map f \FF$.
{{improve|perhaps directly use $p :{{=}} \ds \sup_{S \mathop \in \FF} \inf S$?}}
Then $\FF$ converges to $p$:
=== Upward rays ===
Let $a \in X$ with $a \prec p$.
Since $\FF$ is an ultrafilter, either $Y \cap {\uparr... | Let $\struct {X, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Let $Y \subseteq X$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $X$.
Then $Y$ is a [[Definition:Compact Subspace|compact subspace]] of $\struct {X, \tau}$ {{iff}} both of the following ... | Let $\FF$ be an [[Definition:Ultrafilter on Set|ultrafilter]] on $Y$.
For $S \in \FF$, let $\map f S = \inf S$.
Let $p = \sup \map f \FF$.
{{improve|perhaps directly use $p :{{=}} \ds \sup_{S \mathop \in \FF} \inf S$?}}
Then $\FF$ converges to $p$:
=== Upward rays ===
Let $a \in X$ with $a \prec p$.
Since $\FF$ ... | Compact Subspace of Linearly Ordered Space/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Compact_Subspace_of_Linearly_Ordered_Space | https://proofwiki.org/wiki/Compact_Subspace_of_Linearly_Ordered_Space/Reverse_Implication/Proof_2 | [
"Compact Subspace of Linearly Ordered Space",
"Linearly Ordered Spaces",
"Compact Topological Spaces"
] | [
"Definition:Linearly Ordered Space",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Compact Topological Space/Subspace",
"Definition:Non-Empty Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Non-Empty Set"
] | [
"Definition:Ultrafilter on Set",
"Definition:Ultrafilter on Set",
"Definition:Filter on Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Contradiction",
"Definition:Lower Bound of Set",
"Definition:Infimum of Set",
"Extended Transitivity",
"Definition:Supremum of Set",
... |
proofwiki-7283 | Linearly Ordered Space is Hausdorff | Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Then $\struct {X, \tau}$ is a Hausdorff space. | Let $x, y \in X$ with $x \ne y$.
Since $\le$ is a total ordering, either $x \prec y$ or $y \prec x$.
{{WLOG}}, assume that $x \prec y$.
If there is a $z \in X$ such that $x \prec z \prec y$, then $z^\prec$ and $z^\succ$ separate $x$ and $y$.
Otherwise, by Upper Closure is Strict Upper Closure of Immediate Predecessor, ... | Let $\struct {X, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Then $\struct {X, \tau}$ is a [[Definition:Hausdorff Space|Hausdorff space]]. | Let $x, y \in X$ with $x \ne y$.
Since $\le$ is a [[Definition:Total Ordering|total ordering]], either $x \prec y$ or $y \prec x$.
{{WLOG}}, assume that $x \prec y$.
If there is a $z \in X$ such that $x \prec z \prec y$, then $z^\prec$ and $z^\succ$ separate $x$ and $y$.
Otherwise, by [[Upper Closure is Strict Upp... | Linearly Ordered Space is Hausdorff | https://proofwiki.org/wiki/Linearly_Ordered_Space_is_Hausdorff | https://proofwiki.org/wiki/Linearly_Ordered_Space_is_Hausdorff | [
"Linearly Ordered Spaces",
"Examples of Hausdorff Spaces"
] | [
"Definition:Linearly Ordered Space",
"Definition:T2 Space"
] | [
"Definition:Total Ordering",
"Upper Closure is Strict Upper Closure of Immediate Predecessor",
"Definition:T2 Space",
"Category:Linearly Ordered Spaces",
"Category:Examples of Hausdorff Spaces"
] |
proofwiki-7284 | Subset of Linearly Ordered Space which is Order-Complete and Closed but not Compact | Let $X = \hointr 0 1 \cup \openint 2 3 \cup \set 4$.
Let $\preceq$ be the ordering on $X$ induced by the usual ordering of the real numbers.
Let $\tau$ be the $\preceq$ order topology on $X$.
Let $Y = \hointr 0 1 \cup \set 4$.
Let $\tau'$ be the $\tau$-relative subspace topology on $Y$.
Then:
:$\struct {Y, \preceq}$ is... | First it is demonstrated that $\struct {Y, \preceq}$ is a complete lattice.
Let $\phi: Y \to \closedint 0 1$ be defined as:
:<nowiki>$\map \phi y = \begin{cases}
y & : y \in \hointr 0 1 \\
1 & : y = 4 \end{cases}$</nowiki>
Then $\phi$ is a order isomorphism.
{{explain|The above needs to be proved.}}
{{qed|lemma}}
We ha... | Let $X = \hointr 0 1 \cup \openint 2 3 \cup \set 4$.
Let $\preceq$ be the [[Definition:Ordering|ordering]] on $X$ induced by the [[Definition:Usual Ordering|usual ordering]] of the [[Definition:Real Number|real numbers]].
Let $\tau$ be the $\preceq$ [[Definition:Order Topology|order topology]] on $X$.
Let $Y = \hoin... | First it is demonstrated that $\struct {Y, \preceq}$ is a [[Definition:Complete Lattice|complete lattice]].
Let $\phi: Y \to \closedint 0 1$ be defined as:
:<nowiki>$\map \phi y = \begin{cases}
y & : y \in \hointr 0 1 \\
1 & : y = 4 \end{cases}$</nowiki>
Then $\phi$ is a [[Definition:Order Isomorphism|order isomorphi... | Subset of Linearly Ordered Space which is Order-Complete and Closed but not Compact | https://proofwiki.org/wiki/Subset_of_Linearly_Ordered_Space_which_is_Order-Complete_and_Closed_but_not_Compact | https://proofwiki.org/wiki/Subset_of_Linearly_Ordered_Space_which_is_Order-Complete_and_Closed_but_not_Compact | [
"Order Topologies"
] | [
"Definition:Ordering",
"Definition:Usual Ordering",
"Definition:Real Number",
"Definition:Order Topology",
"Definition:Topological Subspace",
"Definition:Complete Lattice",
"Definition:Closed Set/Topology",
"Definition:Compact Topological Space"
] | [
"Definition:Complete Lattice",
"Definition:Order Isomorphism",
"Definition:Complete Lattice",
"Definition:Closed Set/Topology",
"Definition:Relative Complement",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Compact Topological Space",
"Definition:Lower Closure/Elemen... |
proofwiki-7285 | Rule of Explosion/Variant 2 | :$\vdash \paren {p \land \neg p} \implies q$ | {{BeginTableau|\vdash \paren {p \land \neg p} \implies q}}
{{Assumption|1|p \land \neg p}}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|\neg p|1|2}}
{{Addition|4|1|p \lor q|2|1}}
{{ModusTollendoPonens|5|1|q|4|3}}
{{EndTableau|qed}} | :$\vdash \paren {p \land \neg p} \implies q$ | {{BeginTableau|\vdash \paren {p \land \neg p} \implies q}}
{{Assumption|1|p \land \neg p}}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|\neg p|1|2}}
{{Addition|4|1|p \lor q|2|1}}
{{ModusTollendoPonens|5|1|q|4|3}}
{{EndTableau|qed}} | Rule of Explosion/Variant 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Explosion/Variant_2 | https://proofwiki.org/wiki/Rule_of_Explosion/Variant_2/Proof_1 | [
"Rule of Explosion"
] | [] | [] |
proofwiki-7286 | Jordan Curve Theorem | Let $\gamma: \closedint 0 1 \to \R^2$ be a Jordan curve.
Let $\Img \gamma$ denote the image of $\gamma$.
Then $\R^2 \setminus \Img \gamma$ is a union of two disjoint connected components.
Both components are open in $\R^2$, and both components have $\Img \gamma$ as their boundary.
One component is bounded, and is calle... | {{proof wanted}}
{{Namedfor|Marie Ennemond Camille Jordan|cat = Jordan, C}} | Let $\gamma: \closedint 0 1 \to \R^2$ be a [[Definition:Jordan Curve|Jordan curve]].
Let $\Img \gamma$ denote the [[Definition:Image of Mapping|image]] of $\gamma$.
Then $\R^2 \setminus \Img \gamma$ is a [[Definition:Set Union|union]] of two [[Definition:Disjoint Sets|disjoint]] [[Definition:Connected Topological Sp... | {{proof wanted}}
{{Namedfor|Marie Ennemond Camille Jordan|cat = Jordan, C}} | Jordan Curve Theorem | https://proofwiki.org/wiki/Jordan_Curve_Theorem | https://proofwiki.org/wiki/Jordan_Curve_Theorem | [
"Jordan Curve Theorem",
"Jordan Curves"
] | [
"Definition:Jordan Curve",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Set Union",
"Definition:Disjoint Sets",
"Definition:Connected Topological Space",
"Definition:Component (Topology)",
"Definition:Open Set/Metric Space",
"Definition:Boundary (Topology)",
"Definition:Bounded Metri... | [] |
proofwiki-7287 | Heine-Borel Theorem/Dedekind Complete Space | Let $T = \struct {X, \preceq, \tau}$ be a Dedekind-complete linearly ordered space.
Let $Y$ be a non-empty subset of $X$.
Then $Y$ is compact {{iff}} $Y$ is closed and bounded in $T$. | === Sufficient Condition ===
Let $Y$ be a compact subspace of $T$.
From:
:Linearly Ordered Space is Hausdorff
:Compact Subspace of Hausdorff Space is Closed
it follows that $Y$ is closed in $T$.
From Compact Subspace of Linearly Ordered Space: Lemma 1, $\struct {Y, \preceq {\restriction_Y} }$ is a complete lattice.
{{f... | Let $T = \struct {X, \preceq, \tau}$ be a [[Definition:Dedekind Complete|Dedekind-complete]] [[Definition:Linearly Ordered Space|linearly ordered space]].
Let $Y$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $X$.
Then $Y$ is [[Definition:Compact Subspace|compact]] {{iff}} $Y$ is [[Defi... | === Sufficient Condition ===
Let $Y$ be a [[Definition:Compact Subspace|compact subspace]] of $T$.
From:
:[[Linearly Ordered Space is Hausdorff]]
:[[Compact Subspace of Hausdorff Space is Closed]]
it follows that $Y$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
From [[Compact Subspace of Linearly Ordered... | Heine-Borel Theorem/Dedekind Complete Space | https://proofwiki.org/wiki/Heine-Borel_Theorem/Dedekind_Complete_Space | https://proofwiki.org/wiki/Heine-Borel_Theorem/Dedekind_Complete_Space | [
"Linearly Ordered Spaces",
"Dedekind Completeness Property",
"Heine-Borel Theorem"
] | [
"Definition:Dedekind Completeness Property",
"Definition:Linearly Ordered Space",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Compact Topological Space/Subspace",
"Definition:Closed Set/Topology",
"Definition:Bounded Set"
] | [
"Definition:Compact Topological Space/Subspace",
"Linearly Ordered Space is Hausdorff",
"Compact Subspace of Hausdorff Space is Closed",
"Definition:Closed Set/Topology",
"Compact Subspace of Linearly Ordered Space/Lemma 1",
"Definition:Complete Lattice",
"Definition:Closed Set/Topology",
"Definition:... |
proofwiki-7288 | Heine-Borel iff Dedekind Complete | Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Then $X$ is Dedekind complete {{iff}} every closed, bounded subset of $X$ is compact. | The forward implication follows from Heine-Borel Theorem: Dedekind-Complete Space.
Suppose that $X$ is not Dedekind complete.
Then $X$ has a non-empty subset $S$ with an upper bound $b$ in $X$ but no supremum in $X$.
Let $a \in S$ and let $Y = {\bar \downarrow} S \cap {\bar \uparrow} a$.
$Y$ is nonempty and bounded bel... | Let $\struct {X, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Then $X$ is [[Definition:Dedekind Complete|Dedekind complete]] {{iff}} every closed, bounded subset of $X$ is [[Definition:Compact Subspace|compact]]. | The forward implication follows from [[Heine-Borel Theorem/Dedekind-Complete Space|Heine-Borel Theorem: Dedekind-Complete Space]].
Suppose that $X$ is not Dedekind complete.
Then $X$ has a non-empty subset $S$ with an upper bound $b$ in $X$ but no supremum in $X$.
Let $a \in S$ and let $Y = {\bar \downarrow} S \cap ... | Heine-Borel iff Dedekind Complete | https://proofwiki.org/wiki/Heine-Borel_iff_Dedekind_Complete | https://proofwiki.org/wiki/Heine-Borel_iff_Dedekind_Complete | [
"Order Topologies",
"Compact Topological Spaces"
] | [
"Definition:Linearly Ordered Space",
"Definition:Dedekind Completeness Property",
"Definition:Compact Topological Space/Subspace"
] | [
"Heine-Borel Theorem/Dedekind Complete Space"
] |
proofwiki-7289 | Jordan Polygon Theorem | Let $P$ be a polygon embedded in $\R^2$.
Denote the boundary of $P$ as $\partial P$.
Then, $\R^2 \setminus \partial P$ is a union of two connected components.
Both components are open in $\R^2$.
One component is bounded, and is called the interior of $P$.
The other component is unbounded, and is called the exterior of ... | === Lemma ===
{{:Jordan Polygon Theorem/Lemma 1}}{{qed|lemma}}
We show that $\R^2 \setminus \partial P$ is not path-connected.
Find any $q_1 \in R^2 \setminus \partial P$ and $\theta \in \R$ such that the ray $\LL_\theta = \set {q_1 + s \map g \theta: s \in \R_{\ge 0} }$ has exactly one crossing of $\partial P$.
Find a... | Let $P$ be a [[Definition:Polygon|polygon]] embedded in $\R^2$.
Denote the [[Definition:Boundary (Geometry)|boundary]] of $P$ as $\partial P$.
Then, $\R^2 \setminus \partial P$ is a [[Definition:Set Union|union]] of two [[Definition:Connected Set (Topology)|connected]] [[Definition:Component (Topology)|components]].... | === [[Jordan Polygon Theorem/Lemma 1|Lemma]] ===
{{:Jordan Polygon Theorem/Lemma 1}}{{qed|lemma}}
We show that $\R^2 \setminus \partial P$ is not [[Definition:Path-Connected Metric Subspace|path-connected]].
Find any $q_1 \in R^2 \setminus \partial P$ and $\theta \in \R$ such that the [[Definition:Ray (Geometry)|ray... | Jordan Polygon Theorem | https://proofwiki.org/wiki/Jordan_Polygon_Theorem | https://proofwiki.org/wiki/Jordan_Polygon_Theorem | [
"Jordan Polygon Theorem",
"Topology"
] | [
"Definition:Polygon",
"Definition:Boundary (Geometry)",
"Definition:Set Union",
"Definition:Connected Set (Topology)",
"Definition:Component (Topology)",
"Definition:Open Set/Metric Space",
"Definition:Bounded Metric Space",
"Definition:Jordan Curve/Interior",
"Definition:Bounded Metric Space/Unboun... | [
"Jordan Polygon Theorem/Lemma 1",
"Definition:Path-Connected/Metric Space/Subset",
"Definition:Line/Infinite Half-Line",
"Definition:Crossing (Jordan Curve)",
"Jordan Polygon Parity Lemma",
"Definition:Path (Topology)",
"Definition:Path-Connected/Metric Space/Subset",
"Definition:Set Union",
"Defini... |
proofwiki-7290 | Jordan Polygon Interior and Exterior Criterion | Let $P$ be a polygon embedded in $\R^2$.
Let $q \in \R^2 \setminus \partial P$, where $\partial P$ denotes the boundary of $P$.
Let $\mathbf v \in \R^2 \setminus \set \bszero$ be a non-zero vector.
Let $\LL = \set {q + s \mathbf v: s \in \R_{\ge 0} }$ be a ray with start point $q$.
Let $\map N q \in \N$ be the number o... | From Boundary of Polygon is Jordan Curve, it follows that $\partial P$ is equal to the image of a Jordan curve.
From the Jordan Polygon Theorem, it follows that $\Int P$ and $\Ext P$ are path-connected.
Then, Jordan Polygon Parity Lemma shows that $\map N q = \map {\operatorname{par} } q$, where $\map {\operatorname{pa... | Let $P$ be a [[Definition:Polygon|polygon]] embedded in $\R^2$.
Let $q \in \R^2 \setminus \partial P$, where $\partial P$ denotes the [[Definition:Boundary (Geometry)|boundary]] of $P$.
Let $\mathbf v \in \R^2 \setminus \set \bszero$ be a non-[[Definition:Zero Vector|zero]] [[Definition:Plane Vector|vector]].
Let $\... | From [[Boundary of Polygon is Jordan Curve]], it follows that $\partial P$ is equal to the [[Definition:Image of Mapping|image]] of a [[Definition:Jordan Curve|Jordan curve]].
From the [[Jordan Polygon Theorem]], it follows that $\Int P$ and $\Ext P$ are [[Definition:Path-Connected Set|path-connected]].
Then, [[Jorda... | Jordan Polygon Interior and Exterior Criterion | https://proofwiki.org/wiki/Jordan_Polygon_Interior_and_Exterior_Criterion | https://proofwiki.org/wiki/Jordan_Polygon_Interior_and_Exterior_Criterion | [
"Topology"
] | [
"Definition:Polygon",
"Definition:Boundary (Geometry)",
"Definition:Zero Vector",
"Definition:Vector/Real Euclidean Space/Plane Vector",
"Definition:Line/Infinite Half-Line",
"Definition:Crossing (Jordan Curve)",
"Definition:Jordan Curve/Interior",
"Definition:Jordan Curve/Exterior",
"Definition:Jor... | [
"Boundary of Polygon is Jordan Curve",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Jordan Curve",
"Jordan Polygon Theorem",
"Definition:Path-Connected/Set",
"Jordan Polygon Parity Lemma",
"Definition:Crossing (Jordan Curve)/Parity",
"Jordan Polygon Theorem",
"Definition:Bounded Metr... |
proofwiki-7291 | Closed Set in Linearly Ordered Space | Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $C$ be a subset of $X$.
Then $C$ is closed in $X$ {{iff}} for all non-empty subsets $S$ of $C$:
:If $s \in X$ is a supremum or infimum of $S$ in $X$, then $s \in C$. | === Necessary Condition ===
Suppose that $C$ be closed.
Let $S$ be a non-empty subset of $C$.
Let $b \in X \setminus C$.
We will show that $b$ is not a supremum of $S$.
If $b$ is not an upper bound of $S$, then by definition $b$ cannot be a supremum of $S$.
Suppose, then, that $b$ is an upper bound of $S$.
Since $C$ is... | Let $\struct {X, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Let $C$ be a [[Definition:Subset|subset]] of $X$.
Then $C$ is [[Definition:Closed Set (Topology)|closed]] in $X$ {{iff}} for all [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subsets]] $S$ of $C$:
:If $s ... | === Necessary Condition ===
Suppose that $C$ be [[Definition:Closed Set (Topology)|closed]].
Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $C$.
Let $b \in X \setminus C$.
We will show that $b$ is not a [[Definition:Supremum of Set|supremum]] of $S$.
If $b$ is not an [[Definit... | Closed Set in Linearly Ordered Space | https://proofwiki.org/wiki/Closed_Set_in_Linearly_Ordered_Space | https://proofwiki.org/wiki/Closed_Set_in_Linearly_Ordered_Space | [
"Linearly Ordered Spaces"
] | [
"Definition:Linearly Ordered Space",
"Definition:Subset",
"Definition:Closed Set/Topology",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Supremum of Set",
"Definition:Infimum of Set"
] | [
"Definition:Closed Set/Topology",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Closed Set/Topology",
"Definition:Interval/Ordered Set/Open",
"Definition:Ray... |
proofwiki-7292 | Conditional is Left Distributive over Disjunction/Formulation 1/Forward Implication | :$p \implies \paren {q \lor r} \vdash \paren {p \implies q} \lor \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \lor r} \vdash \paren {p \implies q} \lor \paren {p \implies r} }}
{{Assumption|1|p \implies \paren {q \lor r} }}
{{Assumption|2|p}}
{{ModusPonens|3|1, 2|q \lor r|1|2}}
{{IdentityLaw|4|2|p|2}}
{{Assumption|5|q}}
{{Implication|6|5|p \implies q|4|5}}
{{Addition|7|5|\paren {p \implies ... | :$p \implies \paren {q \lor r} \vdash \paren {p \implies q} \lor \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \lor r} \vdash \paren {p \implies q} \lor \paren {p \implies r} }}
{{Assumption|1|p \implies \paren {q \lor r} }}
{{Assumption|2|p}}
{{ModusPonens|3|1, 2|q \lor r|1|2}}
{{IdentityLaw|4|2|p|2}}
{{Assumption|5|q}}
{{Implication|6|5|p \implies q|4|5}}
{{Addition|7|5|\paren {p \implies ... | Conditional is Left Distributive over Disjunction/Formulation 1/Forward Implication | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Disjunction/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Disjunction/Formulation_1/Forward_Implication | [
"Conditional is Left Distributive over Disjunction"
] | [] | [
"True Statement is implied by Every Statement",
"Category:Conditional is Left Distributive over Disjunction"
] |
proofwiki-7293 | Conditional is Left Distributive over Disjunction/Formulation 1/Reverse Implication | :$\paren {p \implies q} \lor \paren{p \implies r} \vdash p \implies \paren {q \lor r}$ | {{BeginTableau|\paren {p \implies q} \lor \paren {p \implies r} \vdash p \implies \paren {q \lor r} }}
{{Assumption|1|\paren {p \implies q} \lor \paren {p \implies r} }}
{{Assumption|2|p \implies q}}
{{Assumption|3|p}}
{{ModusPonens|4|2, 3|q|2|3}}
{{Addition|5|2, 3|q \lor r|4|1}}
{{Implication|6|2|p \implies \paren {q ... | :$\paren {p \implies q} \lor \paren{p \implies r} \vdash p \implies \paren {q \lor r}$ | {{BeginTableau|\paren {p \implies q} \lor \paren {p \implies r} \vdash p \implies \paren {q \lor r} }}
{{Assumption|1|\paren {p \implies q} \lor \paren {p \implies r} }}
{{Assumption|2|p \implies q}}
{{Assumption|3|p}}
{{ModusPonens|4|2, 3|q|2|3}}
{{Addition|5|2, 3|q \lor r|4|1}}
{{Implication|6|2|p \implies \paren {q ... | Conditional is Left Distributive over Disjunction/Formulation 1/Reverse Implication | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Disjunction/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Disjunction/Formulation_1/Reverse_Implication | [
"Conditional is Left Distributive over Disjunction"
] | [] | [
"Category:Conditional is Left Distributive over Disjunction"
] |
proofwiki-7294 | Factor Principles/Disjunction on Left/Formulation 2 | :$\vdash \paren {p \implies q} \implies \paren {\paren {r \lor p} \implies \paren {r \lor q} }$ | {{BeginTableau|\vdash \paren {p \implies q} \implies \paren {\paren {r \lor p} \implies \paren {r \lor q} } }}
{{Assumption|1|p \implies q}}
{{SequentIntro|2|1|\paren {\paren {r \lor p} \implies \paren {r \lor q} }|1|Factor Principles: Disjunction on Left: Formulation 1}}
{{Implication|3|1|\paren {p \implies q} \implie... | :$\vdash \paren {p \implies q} \implies \paren {\paren {r \lor p} \implies \paren {r \lor q} }$ | {{BeginTableau|\vdash \paren {p \implies q} \implies \paren {\paren {r \lor p} \implies \paren {r \lor q} } }}
{{Assumption|1|p \implies q}}
{{SequentIntro|2|1|\paren {\paren {r \lor p} \implies \paren {r \lor q} }|1|[[Factor Principles/Disjunction on Left/Formulation 1|Factor Principles: Disjunction on Left: Formulati... | Factor Principles/Disjunction on Left/Formulation 2 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Left/Formulation_2 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Left/Formulation_2 | [
"Factor Principles"
] | [] | [
"Factor Principles/Disjunction on Left/Formulation 1"
] |
proofwiki-7295 | Conditional is Left Distributive over Disjunction/Formulation 2 | :$\vdash \paren {p \implies \paren {q \lor r} } \iff \paren {\paren {p \implies q} \lor \paren{p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \lor r} } \iff \paren {\paren {p \implies q} \lor \paren {p \implies r} } }}
{{Assumption|1|p \implies \paren {q \lor r} }}
{{SequentIntro|2|1|\paren {p \implies q} \lor \paren {p \implies r}|1|Conditional is Left Distributive over Disjunction: Formulation 1}}
{{Implic... | :$\vdash \paren {p \implies \paren {q \lor r} } \iff \paren {\paren {p \implies q} \lor \paren{p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \lor r} } \iff \paren {\paren {p \implies q} \lor \paren {p \implies r} } }}
{{Assumption|1|p \implies \paren {q \lor r} }}
{{SequentIntro|2|1|\paren {p \implies q} \lor \paren {p \implies r}|1|[[Conditional is Left Distributive over Disjunction/Formulation 1|Condition... | Conditional is Left Distributive over Disjunction/Formulation 2 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Disjunction/Formulation_2 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Disjunction/Formulation_2 | [
"Conditional is Left Distributive over Disjunction"
] | [] | [
"Conditional is Left Distributive over Disjunction/Formulation 1",
"Conditional is Left Distributive over Disjunction/Formulation 1"
] |
proofwiki-7296 | Factor Principles/Disjunction on Right/Formulation 2 | :$\vdash \paren {p \implies q} \implies \paren {\paren {p \lor r} \implies \paren {q \lor r} }$ | {{BeginTableau|\vdash \paren {p \implies q} \implies \paren {\paren {p \lor r} \implies \paren {q \lor r} } }}
{{Assumption|1|p \implies q}}
{{SequentIntro|2|1|\paren {p \lor r} \implies \paren {q \lor r}|1
|Factor Principles: Disjunction on Right: Formulation 1}}
{{Implication|3|1|\paren {p \implies q} \implies \par... | :$\vdash \paren {p \implies q} \implies \paren {\paren {p \lor r} \implies \paren {q \lor r} }$ | {{BeginTableau|\vdash \paren {p \implies q} \implies \paren {\paren {p \lor r} \implies \paren {q \lor r} } }}
{{Assumption|1|p \implies q}}
{{SequentIntro|2|1|\paren {p \lor r} \implies \paren {q \lor r}|1
|[[Factor Principles/Disjunction on Right/Formulation 1|Factor Principles: Disjunction on Right: Formulation 1]... | Factor Principles/Disjunction on Right/Formulation 2 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Right/Formulation_2 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Right/Formulation_2 | [
"Factor Principles"
] | [] | [
"Factor Principles/Disjunction on Right/Formulation 1"
] |
proofwiki-7297 | Factor Principles/Disjunction on Right/Formulation 1 | :$p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r}$ | {{BeginTableau|p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r} }}
{{Premise|1|p \implies q}}
{{TheoremIntro|2|r \implies r|Law of Identity: Formulation 2}}
{{SequentIntro|3|1|\paren {p \lor r} \implies \paren {q \lor r}|1, 2|Constructive Dilemma}}
{{EndTableau}}
{{qed}} | :$p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r}$ | {{BeginTableau|p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r} }}
{{Premise|1|p \implies q}}
{{TheoremIntro|2|r \implies r|[[Law of Identity/Formulation 2|Law of Identity: Formulation 2]]}}
{{SequentIntro|3|1|\paren {p \lor r} \implies \paren {q \lor r}|1, 2|[[Constructive Dilemma/Formulation 1|Constru... | Factor Principles/Disjunction on Right/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Right/Formulation_1 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Right/Formulation_1/Proof_1 | [
"Factor Principles"
] | [] | [
"Law of Identity/Formulation 2",
"Constructive Dilemma/Formulation 1"
] |
proofwiki-7298 | Factor Principles/Disjunction on Right/Formulation 1 | :$p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r}$ | {{BeginTableau|p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r} }}
{{Premise|1|p \implies q}}
{{Assumption|2|p \lor r}}
{{Assumption|3|r}}
{{Addition|4|3|q \lor r|3|2}}
{{Assumption|5|p}}
{{ModusPonens|6|1, 5|q|1|5}}
{{Addition|7|1, 5|q \lor r|6|1}}
{{ProofByCases|8|1, 2|q \lor r|2|5|7|3|4}}
{{Implicati... | :$p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r}$ | {{BeginTableau|p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r} }}
{{Premise|1|p \implies q}}
{{Assumption|2|p \lor r}}
{{Assumption|3|r}}
{{Addition|4|3|q \lor r|3|2}}
{{Assumption|5|p}}
{{ModusPonens|6|1, 5|q|1|5}}
{{Addition|7|1, 5|q \lor r|6|1}}
{{ProofByCases|8|1, 2|q \lor r|2|5|7|3|4}}
{{Implicati... | Factor Principles/Disjunction on Right/Formulation 1/Proof 2 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Right/Formulation_1 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Right/Formulation_1/Proof_2 | [
"Factor Principles"
] | [] | [] |
proofwiki-7299 | Factor Principles/Disjunction on Right/Formulation 1 | :$p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r}$ | {{BeginTableau|p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r} }}
{{Premise|1|p \implies q}}
{{Assumption|2|p \lor r}}
{{Commutation|3|2|r \lor p|2|Disjunction}}
{{SequentIntro|4|1|\paren {r \lor p} \implies \paren {r \lor q}|1|Factor Principles/Disjunction on Left/Formulation 1}}
{{ModusPonens|5|1,2|r... | :$p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r}$ | {{BeginTableau|p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r} }}
{{Premise|1|p \implies q}}
{{Assumption|2|p \lor r}}
{{Commutation|3|2|r \lor p|2|Disjunction}}
{{SequentIntro|4|1|\paren {r \lor p} \implies \paren {r \lor q}|1|[[Factor Principles/Disjunction on Left/Formulation 1]]}}
{{ModusPonens|5|1... | Factor Principles/Disjunction on Right/Formulation 1/Proof 3 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Right/Formulation_1 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Right/Formulation_1/Proof_3 | [
"Factor Principles"
] | [] | [
"Factor Principles/Disjunction on Left/Formulation 1"
] |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.